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Prelims-H5060.tex 18/7/2007 13: 46 page i

Structural Design of Steelworkto

EN 1993 and EN 1994

Third edition

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Prelims-H5060.tex 18/7/2007 13: 46 page iii

Structural Design ofSteelwork to EN 1993and EN 1994

Third edition

L.H. Martin Bsc, PhD, CEng FICEJ.A. Purkiss Bsc (Eng), PhD

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD

PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO

Butterworth-Heinemann is an imprint of Elsevier

Prelims-H5060.tex 18/7/2007 13: 46 page iv

Butterworth-Heinemann is an imprint of ElsevierLinacre House, Jordan Hill, Oxford OX2 8DP, UK30 Corporate Drive, Suite 400, Burlington, MA 01803, USA

First edition published by Edward Arnold 1984Second edition 1992Third edition 2008

Copyright © 2008, L.H. Martin and J.A. Purkiss. All rights reserved

The right of L.H. Martin and J.A. Purkiss to be identified as the author of this work has been asserted inaccordance with the Copyright, Designs and Patents Act 1988

No part of this publication may be reproduced, stored in a retrieval system or transmitted in any formor by any means electronic, mechanical, photocopying, recording or otherwise without the prior writtenpermission of the publisher

Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK:phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: [email protected]. Alternativelyyou can submit your request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions,and selecting Obtaining permission to use Elsevier material

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

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ISBN13: 978-0-7506-5060-1

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Printed and bound in Great Britan

07 08 09 10 10 9 8 7 6 5 4 3 2 1

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Contents

Preface ixAcknowledgements xiPrincipal Symbols xiii

CHAPTER

1 General 1

1.1 Description of steel structures 11.2 Development, manufacture and types of steel 51.3 Structural design 71.4 Fabrication of steelwork 12

CHAPTER

2 Mechanical Properties of Structural Steel 17

2.1 Variation of material properties 172.2 Characteristic strength 182.3 Design strength 192.4 Other design values for steel 192.5 Corrosion and durability of steelwork 202.6 Brittle fracture 222.7 Residual stresses 222.8 Fatigue 232.9 Stress concentrations 242.10 Failure criteria for steel 24

CHAPTER

3 Actions 27

3.1 Description 273.2 Classification of actions 273.3 Actions varying in time 283.4 Design values of actions 293.5 Actions with spatial variation 31

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vi • Contents

CHAPTER

4 Laterally Restrained Beams 35

4.1 Structural classification of sections 354.2 Elastic section properties and analysis in bending 384.3 Elastic shear stresses 544.4 Elastic torsional shear stresses 624.5 Plastic section properties and analysis 674.6 Effect of shear force on the plastic moment of resistance 734.7 Lateral restraint 774.8 Resistance of beams to transverse forces 78

CHAPTER

5 Laterally Unrestrained Beams 91

5.1 Lateral torsional buckling of rolled sections symmetric about both axes 915.2 Pure torsional buckling 1275.3 Plate girders 132

CHAPTER

6 Axially Loaded Members 175

6.1 Axially loaded tension members 1756.2 Combined bending and axial force – excluding buckling 1776.3 Buckling of axially loaded compression members 1806.4 Combined bending and axial force – with buckling 190

CHAPTER

7 Structural Joints 198

7.1 Introduction 1987.2 The ideal structural joint 1997.3 Welded joints 1997.4 Bolted joints 2077.5 Plate thicknesses for joint components 2187.6 Joints subject to shear forces 2247.7 Joints subject to eccentric shear forces 2257.8 Joints with end bearing 2267.9 ‘Pinned’ joints 2287.10 ‘Rigid’ joints 230

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Contents • vii

7.11 Joint rotational stiffness 2757.12 Frame-to-joint stiffness 277

CHAPTER

8 Frames and Framing 282

8.1 Single storey structures 2828.2 Multi-storey construction 2838.3 Influence of connection design and detailing 2868.4 Structural actions 2868.5 Single storey structures under horizontal loading 2888.6 Multi-storey construction 2938.7 Behaviour under accidental effects 2948.8 Transmission of loading 2988.9 Design of bracing 3008.10 Fire performance 3018.11 Additional design constraints 3038.12 Design philosophies 3068.13 Design issues for multi-storey structures 3108.14 Portal frame design 314

CHAPTER

9 Trusses 341

9.1 Triangulated trusses 3419.2 Non-triangulated trusses 347

CHAPTER

10 Composite Construction 358

10.1 Composite slabs 35810.2 Design of decking 35910.3 Composite beams 37010.4 Composite columns 395

CHAPTER

11 Cold-formed Steel Sections 413

11.1 Analytical model 41411.2 Local buckling 419

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viii • Contents

11.3 Distortional buckling 42411.4 Lateral–torsional buckling 43011.5 Calculation of deflections 43411.6 Finite strip methods 43511.7 Design methods for beams partially restrained by sheeting 44211.8 Working examples 44511.9 Chapter conclusions 453

ANNEX 458

INDEX 469

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Preface

This book conforms to the latest recommendations for the design of steel and com-posite steel–concrete structures as described in Eurocode 3: Design of steel structuresand Eurocode 4: Design of composite steel–concrete structures. References to rele-vant clauses of the Codes are given where appropriate. Note that for normal steelworkdesign, including joints, three sections of EN 1993 are required:

• Part 1–1 General rules and rules for buildings• Part 1–5 Plated structural elements• Part 1–8 Design of joints

Additionally if design for cold formed sections is carried out from first principles thenPart 1–3 Cold formed thin gauge members and sheeting is also required.

Whilst it has not been assumed that the reader has a knowledge of structural design,a knowledge of structural mechanics and stress analysis is a prerequisite. However,as noted below certain specialist areas of analysis have been covered in detail sincethe Codes do not provide the requisite information. Thus the book contains detailedexplanations of the principles underlying steelwork design and provides appropriatereferences and suggestions for further reading.

The text should prove useful to students reading for engineering degrees at University,especially for design projects. It will also aid designers who require an introduction tothe new Eurocodes.

For those familiar with current practice, the major changes are:

(1) There is need to refer to more than one part of the various codes with calculationsgenerally becoming more extensive and complex.

(2) Steelwork design stresses are increased as the gamma values on steel are taken as1,0, and the strength of high yield reinforcement is 500 MPa albeit with a gammafactor of 1,15.

(3) A deeper understanding of buckling phenomena is required as the Codes do notsupply the relevant formulae.

(4) Flexure and axial force interaction equations are more complex, thus increasingthe calculations for column design.

(5) The checking of webs for in-plane forces is more complex.(6) Although tension field theory (or its equivalent) may be used for plate girders,

the calculations are simplified compared to earlier versions of the Code.(7) Joints are required to be designed for both strength and stiffness.(8) More comprehensive information is given on thin-walled sections.

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Acknowledgements

Lawrence Martin and John Purkiss would like to thank Long-yuan Li and Xiao-tingChu for writing Chapter 11 on Cold-formed steel sections.

The second author would like to thank Andrew Orton (Corus) for help with problemsover limiting critical lengths for lateral–torsional buckling of rolled sections.

The authors further wish to thank the following for permission to reproduce material:

• Albion Sections Ltd for Fig. 11.1• www.access-steel.com for Figs 11.24 and 11.25• Karoly Zalka and the Institution of Civil Engineers for Fig. 8.13• BSI

BSI Ref. Book Ref.

BS 5950: Part 1: 1990 Tables 15–17 Annex A7

BS 5950: Part 1: 2000 Table 14 Table 5.3

EN 1993-1-3 Fig. 5.6 Fig. 11.8Fig. 5.7a Fig. 11.9Fig. 10.2 Fig. 11.20

EN 1994-1-1 Fig. 9.8 Fig. 10.3

British Standards may be obtained from BSI Customer Services, 389 Chiswick HighRoad, London W4 4AL. Tel: +44 (0)20 8996 9001. e-mail: [email protected]

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Principal Symbols

Listed below are the symbols and suffixes common to European Codes

LATIN UPPER AND LOWER CASEA accidental action; areaa distance; throat thickness of a weldB bolt forceb breadthc outstandd depth of web; diameterE modulus of elasticitye edge distance; end distanceF action; forcef strength of a materialG permanent action; shear modulus of steelH total horizontal load or reaction; warping constant of sectionh heighti radius of gyrationI second moment of areak stiffnessL length; span; buckling lengthl effective buckling length; torsion constant; warping constantM bending momentN axial forcen numberp pitch; spacingQ variable action; prying forceq uniformly distributed actionR resistance; reactionr radius; root radius; number of redundanciesS stiffnesss staggered pitch; distance; bearing lengthT torsional momentt thicknessuu principal major axisvv principal minor axis

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xiv • Principal Symbols

V shear force; total vertical load or reactionv shear stressW section modulusw deflection

GREEK LOWER CASEα coefficient of linear thermal expansion; angle; ratio; factorβ angle; ratio; factorγ partial safety factorδ deflection; deformationε strain; coefficient (235/fy)1/2 where fy is in MPaη distribution factor; shear area factor; critical buckling mode; buckling

imperfection coefficientθ angle; slopeλ slenderness ratio; ratioμ slip factorν Poisson’s ratioρ unit mass; factorσ normal stress; standard deviationτ shear stressφ rotation; slope; ratioχ reduction factor for bucklingψ stress ratio

SUFFIXESEd design strengthel elasticf flangej jointo initial; holep platepl plasticRd resistance strengtht torsionu ultimate strengthv shearw web; warpingx x-x axisy y-y axis; yield strengthz z-z axis

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C h a p t e r 1 / General

1.1 DESCRIPTION OF STEEL STRUCTURES

1.1.1 Shapes of Steel Structures

The introduction of structural steel, circa 1856, provided an additional building mater-ial to stone, brick, timber, wrought iron and cast iron. The advantages of steel are highstrength, high stiffness and good ductility combined with relative ease of fabricationand competitive cost. Steel is most often used for structures where loads and spansare large and therefore is not often used for domestic architecture.

Steel structures include low-rise and high-rise buildings, bridges, towers, pylons, floors,oil rigs, etc. and are essentially composed of frames which support the self-weight,dead loads and external imposed loads (wind, snow, traffic, etc.). For convenienceload bearing frames may be classified as:

(a) Miscellaneous isolated simple structural elements (e.g. beams and columns) orsimple groups of elements (e.g. floors).

(b) Bridgeworks.(c) Single storey factory units (e.g. portal frames).(d) Multi-storey units (e.g. tower blocks).(e) Oil rigs.

A real structure consists of a load bearing frame, cladding and services as shown inFig. 1.1(a). A load bearing frame is an assemblage of members (structural elements)arranged in a regular geometrical pattern in such a way that they interact through struc-tural connections to support loads and maintain them in equilibrium without excessivedeformation. Large deflections and distortions in structures are controlled by the useof bracing which stiffens the structure and can be in the form of diagonal structuralelements, masonry walls, reinforced concrete lift shafts, etc. A load bearing steel frameis idealized, for the purposes of structural design, as center lines representing struc-tural elements which intersect at joints, as shown in Fig. 1.1(b). Other shapes of loadbearing frames are shown in Figs 1.1 (c) to (e).

Structural elements are required to resist forces and displacements in a variety of ways,and may act in tension, compression, flexure, shear, torsion or in any combination of

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2 • Chapter 1 / General

(a) Real structure (b) Idealized load bearing frame

Bracing

(c) (d)

(e)

Pinnedconnection

Dead and snow loading

Windloading

Services

Cladding

Connection

Load bearingframe

Connection

Rigid connection

FIGURE 1.1 Typical load bearing frames

these forces. The structural behaviour of a steel element depends on the nature of theforces, the length and shape of the cross section of the member, the elastic propertiesand the magnitude of the yield stress. For example a tie behaves in a linear elasticmanner until yield is reached. A slender strut behaves in a non-linear elastic manneruntil first yield is attained, provided that local buckling does not occur first. A laterallysupported beam behaves elastically until a plastic hinge forms, while an unbracedbeam fails by elastic torsional buckling. These modes of behaviour are considered indetail in the following chapters.

The structural elements are made to act as a frame by connections. These are com-posed of plates, welds and bolts which are arranged to resist the forces involved. Theconnections are described for structural design purposes as pinned, semi-rigid andrigid, depending on the amount of rotation, and are described, analysed and designedin detail in Chapter 7.

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Structural Design of Steelwork to EN 1993 and EN 1994 • 3

1.1.2 Standard Steel Sections

The optimization of costs in steel construction favours the use of structural steel elem-ents with standard cross-sections and common bar lengths of 12 or 15 m. The billetsof steel are hot rolled to form bars, flats, plates, angles, tees, channels, I sections andhollow sections as shown in Fig. 1.2. The detailed dimensions of these sections aregiven in BS 4, Pt 1 (2005), BSEN 10056-1 (1990), and BSEN 10210-2 (1997).

Where thickness varies, for example, Universal beams, columns and channels, sec-tions are identified by the nominal size, that is, ‘depth × breadth × mass per unitlength × shape’. Where thickness is constant, for example, tees and angle sections,the identification is ‘breadth × depth × thickness × shape’. In addition a section isidentified by the grade of steel.

To optimize on costs steel plates should be selected from available stock sizes. Thick-nesses are in the range of 6, 8, 10, 12,5, 15 mm and then in 5 mm increments.Thicknesses of less than 6 mm are available but because of lower strength and poorercorrosion resistance their use is limited to cold formed sections. Stock plate widthsare in the range 1, 1,25, 1,5, 2, 2,5 and 3 m, but narrow plate widths are also available.Stock plate lengths are in the range 2, 2,5, 3, 4, 5, 6, 10 and 12 m. The adoption ofstock widths and lengths avoids work in cutting to size and also reduces waste.

The application of some types of section is obvious, for example, when a member is intension a round or flat bar is the obvious choice. However, a member in tension may

Universalbeam (UB)

Universalcolumn (UC)

Channel Angle Structural teefrom UB

Circularhollow section

Retangularhollow section

Bars Plate

FIGURE 1.2 Standard steel sections

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be in compression under alternative loading and an angle, tee, or tube is often moreappropriate. The connection at the end of a bar or tube, however, is more difficultto make.

If a structural element is in bending about one axis then the ‘I’ section is the mostefficient because a large proportion of the material is in the flanges, that is, at theextreme fibres. Alternatively, if a member is in bending about two axes at right anglesand also supports an axial load then a tube, or rectangular hollow section, is moreappropriate.

Other steel sections available are cold formed from steel plate into a variety of crosssections for use as lightweight lattice beams, glazing bars, shelf racks, etc. Not allthese sections are standardized because of the large variety of possible shapes anduses, however, there is a wide range of sections listed in BSEN 10162 (2003). Localbuckling can be a problem and edges are stiffened using lips. Also when used asbeams the relative thinness of the material may lead to web crushing, shear bucklingand lateral torsional buckling. Although the thickness of the material (1–3 mm) is lessthan that of the standard sections the resistance to corrosion is good because of thesurface finish obtained by pickling and oiling. After degreasing this surface can beprotected by galvanizing, or painting, or plastic coating. The use in building of coldformed sections in light gauge plate, sheet and strip steel 6 mm thick and under is dealtwith in BSEN 5950 (2001) and EN 1993-1-1 (2005).

1.1.3 Structural Classification of Steel Sections(cl 5.5. EN 1993-1-1 (2005))

A section, or element of a member, in compression due to an axial load may fail bylocal buckling. Local buckling can be avoided by limiting the width to thickness ratios(b/tf or d/tw) of each element of a cross-section. The use of the limiting values givenin Table 5.2, EN 1993-1-1 (2005) avoids tedious and complicated calculations.

Depending on the b/tf or d/tw ratios standard or built-up sections are classified forstructural purposes as:

• Class 1: Low values of b/tf or d/tw where a plastic hinge can be developedwith sufficient rotation capacity to allow redistribution of moments within thestructure.

• Class 2: Full plastic moment capacity can be developed but local buckling mayprevent development of a plastic hinge with sufficient rotation capacity to permitplastic design.

• Class 3: High values of b/tf and d/tw, where stress at the extreme fibres can reachdesign strength but local buckling may prevent the development of the full plasticmoment.

• Class 4: Local buckling may prevent the stress from reaching the design strength.Effective widths are used to allow for local buckling (cl 5.5.2(2), EN 1993-1-1 (2005)).

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1.1.4 Structural Joints (EN 1993-1-8 (2005))

Structural elements are connected together at joints which are not necessarily at theends of members. A structural connection is an assembly of components (plates, bolts,welds, etc.) arranged to transmit forces from one member to another. A connectionmay be subject to any combination of axial force, shear force and bending momentin relation to three perpendicular axes, but for simplicity, where appropriate, thesituation is reduced to forces in one plane.

There are other types of joints in structures which are not structural connections. Forexample a movement joint is introduced into a structure to take up the free expan-sion and contraction that may occur on either side of the joint due to temperature,shrinkage, expansion, creep, settlement, etc. These joints may be detailed to be water-tight but do not generally transmit forces. Detailed recommendations are given byAlexander and Lawson (1981). Another example is a construction joint which is intro-duced because components are manufactured to a convenient size for transportationand need to be connected together on site. In some cases these joints transmit forcesbut in other situations may only need to be waterproof.

1.2 DEVELOPMENT, MANUFACTURE AND TYPES OF STEEL

1.2.1 Outline of Developments in Design Using Ferrous Metals

Prior to 1779, when the Iron Bridge at Coalbrookdale on the Severn was completed, themost important materials used for load bearing structures were masonry and timber.Ferrous materials were only used for fastenings, armaments and chains.

The earliest use of cast iron columns in factory buildings (circa 1780) enabled relativelylarge span floors to be constructed. Due to a large number of disastrous fires around1795, timber beams were replaced by cast iron with the floors carried on brick jackarches between the beams. This mode of construction was pioneered by Strutt in aneffort to attain a fire proof construction technique.

Cast iron, however, is weak in tension and necessitates a tension flange larger thanthe compression flange and consequently cast iron was used mainly for compressionmembers. Large span cast iron beams were impractical, and on occasions disastrousas in the collapse of the Dee bridge designed by Robert Stephenson in 1874. The lastprobable use of cast iron in bridge works was in the piers of the Tay bridge in 1879 whenthe bridge collapsed in high winds due to poor design and unsatisfactory supervisionduring construction.

In an effort to overcome the tensile weakness of cast iron, wrought iron was introducedin 1784 by Henry Cort. Wrought iron enabled the Victorian engineers to producethe following classic structures. Robert Stephenson’s Brittania Bridge was the firstbox girder bridge and represented the first major collaboration between engineer,fabricator (Fairburn) and scientist (Hodgkinson). I.K. Brunel’s Royal Albert Bridge

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at Saltash combined an arch and suspension bridge. Telford’s Menai suspension bridgeused wrought iron chains which have sine been replaced by steel chains. Telford’s PontCysyllte is a canal aqueduct near Llangollen. The first of the four structures wasreplaced after a fire in 1970. The introduction of wrought iron revolutionized shipbuilding and enabled Brunel to produce the S.S. Great Britain.

Steel was first produced in 1740, but was not available in large quantities until Bessemerinvented the converter in 1856. The first major structure to use the new steel exclusivelywas Fowler and Baker’s railway bridge at the Firth of Forth. The first steel rail wasrolled in 1857 and installed at Derby where it was still in use 10 years later. Cast ironrails in the same position lasted about 3 months. Steel rails were in regular productionat Crewe under Ramsbottom from 1866.

By 1840 standard shapes in wrought iron, mainly rolled flats, tees and angles, werein regular production and were appearing in structures about 10 years later. Com-pound girders were fabricated by riveting together the standard sections. Wroughtiron remained in use until around the end of the nineteenth century.

By 1880 the rolling of steel ‘I’ sections had become widespread under the influenceof companies such as Dorman Long. Riveting continued in use as a fastening methoduntil around 1950 when it was superseded by welding. Bessemer steel productionin Britain ended in 1974 and the last open hearth furnace closed in 1980. Furtherinformation on the history of steel making can be found in Buchanan (1972), Cossons(1975), Derry and Williams (1960), Pannel (1964) and Rolt (1970).

1.2.2 Manufacture of Steel Sections

The manufacture of standard steel sections, although now a continuous process, canbe conveniently divided into three stages:

(1) Iron production(2) Steel production(3) Rolling.

Iron production is a continuous process and consists of chemically reducing iron orein a blast furnace using coke and crushed limestone. The resulting material, calledcast iron, is high in carbon, sulphur and phosphorus.

Steel production is a batch process and consists in reducing the carbon, sulphur andphosphorus levels and adding, where necessary, manganese, chromium, nickel, van-adium, etc. This process is now carried out using a Basic Oxygen Converter, whichconsists of a vessel charged with molten cast iron, scrap steel and limestone throughwhich oxygen is passed under pressure to reduce the carbon content by oxidation.This is a batch process which typically produces about 250–300 tons every 40 min. Thealternative electric arc furnace is in limited use (approximately 5% of the UK steelproduction), and is generally used for special steels such as stainless steel.

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From the converter the steel is ‘teemed’ into ingots which are then passed to the rollingmills for successive reduction in size until the finished standard section is produced.The greater the reduction in size the greater the work hardening, which produces vary-ing properties in a section. The variation in cooling rates of different thicknesses intro-duces residual stresses which may be relieved by the subsequent straightening process.Steel plate is now produced using a continuous casting procedure which eliminates,ingot casting, mould stripping, heating in soaking pits and primary rolling. Continuouscasting permits, tighter control, improved quality, reduced wastage and lower costs.

1.2.3 Types of Steel

The steel used in structural engineering is a compound of approximately 98% ironand small percentages of carbon, silicon, manganese, phosphorus, sulphur, niobiumand vanadium as specified in BS 4360 (1990). Increasing the carbon content increasesstrength and hardness but reduces ductility and toughness. Carbon content thereforeis restricted to between 0,25% and 0,2% to produce a steel that is weldable and notbrittle. The niobium and vanadium are introduced to raise the yield strength of thesteel; the manganese improves corrosion resistance; and the phosphorus and sulphurare impurities. BS 4360 (1990) also specifies tolerances, testing procedure and specificrequirements for weldable structural steel.

Steels used in practice are identified by letters and number, for example, S235 is steelwith a tensile yield strength of 235 MPa (Table 3.1, EN 1993-1-1 (2005)).

1.3 STRUCTURAL DESIGN

1.3.1 Initiation of a Design

The demand for a structure originates with the client. The client may be a privateperson, private or public firm, local or national government, or a nationalized industry.

In the first stage preliminary drawings and estimates of costs are produced, followedby consideration of which structural materials to use, that is, reinforced concrete,steel, timber, brickwork, etc. If the structure is a building, an architect only may beinvolved at this stage, but if the structure is a bridge or industrial building then a civilor structural engineer prepares the documents.

If the client is satisfied with the layout and estimated costs then detailed design calcula-tions, drawings and costs are prepared and incorporated in a legal contract document.The design documents should be adequate to detail, fabricate and erect the structure.

The contract document is usually prepared by the consultant engineer and work iscarried out by a contractor who is supervised by the consultant engineer. However,larger firms, local and national government, and nationalized industries, generallyemploy their own consultant engineer.

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The work is generally carried out by a contractor, but alternatively direct labourmay be used. A further alternative is for the contractor to produce a design andconstruct package, where the contractor is responsible for all parts and stages ofthe work.

1.3.2 The Object of Structural Design

The object of structural design is to produce a structure that will not become unser-viceable or collapse in its lifetime, and which fulfils the requirements of the client anduser at reasonable cost.

The requirements of the client and user may include any or all of the following:

(a) The structure should not collapse locally or overall.(b) It should not be so flexible that deformations under load are unsightly or alarm-

ing, or cause damage to the internal partitions and fixtures; neither should anymovement due to live loads, such as wind, cause discomfort or alarm to theoccupants/users.

(c) It should not require excessive repair or maintenance due to accidental overload,or because of the action of weather.

(d) In the case of a building, the structure should be sufficiently fire resistant to, givethe occupants time to escape, enable the fire brigade to fight the fire in safety andto restrict the spread of fire to adjacent structures.

The designer should be conscious of the costs involved which include:

(a) The initial cost which includes fees, site preparation, cost of materials andconstruction.

(b) Maintenace costs (e.g. decoration and structural repair).(c) Insurance chiefly against fire damage.(d) Eventual demolition.

It is the responsibility of the structural engineer to design a structure that is safe andwhich conforms to the requirements of the local bye-laws and building regulations.Information and methods of design are obtained from Standards and Codes of Prac-tice and these are ‘deemed to satisfy’ the local bye-laws and building regulations. Inexceptional circumstances, for example, the use of methods validated by research ortesting, an alternative design may be accepted.

A structural engineer is expected to keep up to date with the latest research informa-tion. In the event of a collapse or malfunction where it can be shown that the engineerhas failed to reasonably anticipate the cause or action leading to collapse, or has failedto apply properly the information at his disposal, that is, Codes of Practice, BritishStandards, Building Regulations, research or information supplied by the manufac-turers, then he may be sued for professional negligence. Consultants and contractorscarry liability insurance to mitigate the effects of such legal action.

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1.3.3 Limit State Design (cl 2.2, EN 1993-1-1 (2005))

It is self-evident that a structure should be ‘safe’ during its lifetime, that is, free fromthe risk of collapse. There are, however, other risks associated with a structure and theterm safe is now replaced by the term ‘serviceable’. A structure should not during itslifetime become ‘unserviceable’, that is, it should be free from risk of collapse, rapiddeterioration, fire, cracking, excessive deflection, etc.

Ideally it should be possible to calculate mathematically the risk involved in struc-tural safety based on the variation in strengths of the material and variation in theloads. Reports, such as the CIRIA Report 63 (1977), have introduced the designerto elegant and powerful concept of ‘structural reliability’. Methods have been devisedwhereby engineering judgement and experience can be combined with statistical analy-sis for the rational computation of partial safety factors in codes of practice. However,in the absence of complete understanding and data concerning aspects of structuralbehaviour, absolute values of reliability cannot be determined.

It is not practical, nor is it economically possible, to design a structure that will neverfail. It is always possible that the structure will contain material that is less than therequired strength or that it will be subject to loads greater than the design loads. Ifactions (forces) and resistance (strength of materials) are determined statistically thenthe relationship can be represented as shown in Fig. 1.3. The design value of resistance(Rd) must be greater than the design value of the actions (Ad).

It is therefore accepted that 5% of the material in a structure is below the designstrength, and that 5% of the applied loads are greater than the design loads. This doesnot mean therefore that collapse is inevitable, because it is extremely unlikely that theweak material and overloading will combine simultaneously to produce collapse.

The philosophy and objectives must be translated into a tangible form using calcula-tions. A structure should be designed to be safe under all conditions of its useful life

Actions A

Frequency

Resistance R

Ad Rd

Rd − Ad > 0Design value of actions Design value of resistance

Resistance or actions

FIGURE 1.3 Statistical relationship between actions and resistance

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and to ensure that this is accomplished certain distinct performance requirements,called ‘limit states’, have been identified. The method of limit state design recognizesthe variability of loads, materials, construction methods and approximations in thetheory and calculations.

Limit states may be at any stage of the life of a structure, or at any stage of loadingand are important for the design of steelwork. To reduce the number of load cases tobe considered only serviceability and ultimate limit states are specified. Each of thesesections is subdivided although some may not be critical in every design. Calculationsfor limit states involve loads and load factors (Chapter 3), and material factors andstrengths (Chapter 2).

Stability, an ultimate limit state, is the ability of a structure, or part of a structure,to resist overturning, overall failure and sway. Calculations should consider the worstrealistic combination of loads at all stages of construction.

All structures, and parts of structures, should be capable of resisting sway forces,for example, by the use of bracing, ‘rigid’ joints, or shear walls. Sway forces arisefrom horizontal loads, for example, winds, and also from practical imperfections, forexample, lack of verticality. The sway forces from practical imperfections are difficultto quantify and advice is given in cl 5.3.3, EN 1993-1-1 (2005).

Also involved in limit state design is the concept of structural integrity. Essentially thismeans that the structure should be tied together as a whole, but if damage occurs, itshould be localized.

Deflection is a serviceability limit state. Deflections should not impair the efficiencyof a structure, or its components, nor cause damage to the finishes. Generally theworst realistic combination of unfactored imposed loads is used to calculate elasticdeflections. These values are compared with empirical values related to the length ofa member or height.

Dynamic effects to be considered at the serviceability limit state are vibrations causedby machines, and oscillations caused by harmonic resonance, for example, wind gustson buildings. The natural frequency of the building should be different from theexciting source to avoid resonance.

Fortunately there are few structural failures and when they do occur they are oftenassociated with human error involved in design calculations, or construction, or in theuse of the structure.

1.3.4 Structural Systems

Structural frame systems may be described as:

(a) simple frames,(b) continuous frames,(c) semi-continuous frames.

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These titles refer to the types of joints and whether bracing is included.

Simple design assumes that ‘pin joints’ connect the members and joint rotations areprevented by bracing. Historically this method was popular because parts of the struc-ture could be designed in isolation and calculations could be done by hand. With theadvent of the computer calculations are less onerous but the method is still in use.

Continuous frames assume that the connections between members are rigid and there-fore the angles between members can be maintained without the use of bracing.Calculations for the design of members and connections are more complicated and acomputer is generally used. Global analysis of the frame is based on elastic, plastic, orelastic–plastic analysis assuming full continuity.

Semi-continuous frames acknowledges that in reality, end moments and rotationsexist at the connections. Global analysis using the computer is based on the moment–rotation and force displacement characteristics of the connections. Bracing is oftennecessary for this type of frame to reduce sway.

1.3.5 Errors

The consequences of an error in structural design can lead to loss of life and damageto property, and it is necessary to appreciate where errors can occur. Small errorsin design calculations can occur in the rounding off of figures but these generally donot lead to failures. The common sense advice is that the accuracy of the calculationshould match the accuracy of the values given in the European Code.

Errors that occur in structural design calculations and which affect structuralsafety are:

(1) Ignorance of the physical behaviour of the structure under load and which con-sequently introduces errors in the basic assumptions used in the theoreticalanalysis.

(2) Errors in estimating the loads, especially the erection forces.(3) Numerical errors in the calculations. These should be eliminated by checking, but

when speed is paramount checks are often ignored.(4) Ignorance of the significance of certain effects (e.g. residual stresses, fatigue, etc).(5) Introduction of new materials, or methods, which have not been proved by tests.(6) Insufficient allowance for tolerances or temperature strains.(7) Insufficient information (e.g. in erection procedures).

Errors that can occur in workshops or on construction sites are:

(1) Using the wrong grade of steel, and when welding using the wrong type ofelectrode.

(2) Using the wrong weight of section. A number of sections are the same nominalsize but differ in web or flange thickness.

(3) Errors in manufacture (e.g. holes in the wrong position).

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Errors that occur in the life of a structure and also affect safety are:

(1) Overloading(2) Removal of structural material (e.g. to insert service ducts)(3) Poor maintenance.

1.4 FABRICATION OF STEELWORK

1.4.1 Drawings

Detailed design calculations are essential for any steel work design but the sizes of themembers, dimensions and geometrical arrangement are usually presented as drawings.Initially the drawings are used by the fabricator and eventually by the contractor onsite. General arrangement drawings are often drawn to scale of 1:100, while detailsare drawn to a scale of 1:20 or 1:10. Special details are drawn to larger scales wherenecessary.

Drawings should be easy to read and should not include superfluous detail. Someimportant notes are:

(a) Members and components should be identified by logically related mark numbers,for example, related to the grid system used in the drawings.

(b) The main members should be presented by a bold outline (0,4 mm wide) anddimension lines should be unobtrusive (0,1 mm wide).

(c) Dimensions should be related to centre lines, or from one end; strings of dimen-sions should be avoided. Dimensions should appear once only so that ambiguitycannot arise when revisions occur. Fabricators should not be put in the positionof having to do arithmetic in order to obtain an essential dimension.

(d) Tolerances for erection purposes should be clearly shown.(e) The grade of steel to be used should be clearly indicated.(f) The size, weight and type of section to be used should be clearly stated.(g) Detailing should take account of possible variations due to rolling margins and

fabrication variations.(h) Keep the design and construction as simple as possible. Where possible use simple

connections, avoid stiffeners, use the minimum number of sections and avoidchanges in section along the length of a member.

(i) Site access, transport and use of cranes should be considered.

1.4.2 Tolerances (cl 3.2.5, EN 1993-1-1 (2005))

Tolerances are limits places on unintentional inaccuracies that occur in dimensionswhich must be allowed for in design if structural elements and components are to fittogether. In steelwork variations occur in the rolling process, marking out, cutting anddrilling during fabrication, and in setting out during erection.

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In the rolling process the allowable tolerances for length, width, thickness and flat-ness for plates are given in BS 4360 (1990). Length and width tolerances are positivewhile those for thickness and flatness are negative and positive. The dimensional andweight tolerances for sections are given in BS 4 Pt 1 (2005), or BS 10056-1 (1990), asappropriate.

During fabrication there is a tendency for members and components to increase ratherthan reduce, and the tolerance is therefore often specified as negative; it is oftencheaper and simpler to insert packing rather than shorten a member, provided that thepacking is not excessive. Where concrete work is associated with steelwork variationsin dimensions are likely to be greater. When casting concrete, for example, errors indimensions may arise from shrinkage or from warping of the shuttering, especiallywhen it is re-used. Therefore, by virtue of the construction method, larger tolerancesare specified for work involving concrete.

To facilitate erection all members and connections should be provided with the max-imum tolerance that is acceptable from structural and architectural considerations.A typical example is a connection between a steel column and a reinforced concretebase. It would produce great difficulties if the base were set too high and a toleranceof approximately 50 mm is often included in the design, with provision for groutingunder the base. Tolerances are also provided to allow lateral adjustment of the foun-dation bolts. Tolerances between concrete and steelwork are also important becausetwo different contractors are involved.

1.4.3 Fabrication, Assembly and Erection of Steelwork

The drawings produced by the structural designer are used first by the steel fabricatorand later by the contractor on site.

The steel fabricator obtains the steel either direct from the rolling mills or from thesteel stockiest, and then cuts, drills and welds the steel components to form the struc-tural elements as shown on the drawings. In general, for British practice, the weldingis confined to the workshop and the connections on site are made using bolts. InAmerican, however, site welding is common practice.

When marking out, the measurements of length for overall size, position of holes, etc.can be done by hand, but if there are several identical components then wooden orcardboard templates are made and repeated measurements avoided. Now automaticmachines, controlled by a computer, or punched paper tape, are used to cut and drillstandard sections. When completed, the steel work should be marked clearly andmanufactured to the accepted tolerances.

When fabricated, parts of the structure are delivered to the site in the largest piecesthat can be transported and erected. For example a lattice girder may be sent fullyassembled to a site in this country, but sent in pieces to fit a standard transport container

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for erection abroad. All components should be assembled within the tolerances andcambers specified, and should not be bent or twisted or otherwise damaged.

On site the general contractor may be responsible for the assembly, erection, con-nections, alignment and leveling of the complete structure. Alternatively the erectionwork may be done by the steel fabricator, or sublet to a specialist steel erector. Theobjective of the erection process is to assemble the steelwork in the most cost-effectivemethod whilst maintaining the stability of individual members, and/or part or completestructure. To do this it may be necessary to introduce cranes and temporary bracingwhich must also be designed to resist the loads involved.

During assembly on site it is inevitable that some components will not fit, despite thetolerances that have been allowed. A typical example is that the faying surfaces for afriction grip joint are not in contact when the bolts are stressed. Other examples aregiven by Mann and Morris (1981). The correction of some faults and the consequentlitigation can be expensive.

1.4.4 Testing of Steelwork (cl 2.5, EN 1993-1-1 (2005))

Steel is routinely sampled and tested during production to maintain quality. However,occasionally new methods of construction are suggested and there may be some doubtas to the validity of the assumptions of behaviour of the structure. Alternatively if thestructure collapses there may be some dispute as to the strength of a component, ormember, of the structure. In such cases testing of components, or part of the structure,may be necessary. However it is generally expensive because of, the accuracy required,cost of material, cost of fabrication, necessity to repeat tests to allow for variationsand to report accurately.

Tests may be classified as:

(a) acceptance tests – non-destructive for confirming structural performance,(b) strength tests – used to confirm the calculated capacity of a component or

structure,(c) tests to failure – to determine the real mode of failure and the true capacity of a

specimen,(d) check tests – where the component assembly is designed on the basis of tests.

The size, shape, position of the gauges, and method of testing of small sample piecesof steel is given in BS 4360 (1990) and BSEN 10002-1 (2001). The tensile test is mostfrequently employed, and gives values of, Young’s modulus, limit of proportionality,yield stress or proof stress, percentage elongation and ultimate stress. Methods ofdestructive testing fusion welded joints and weld metal in steel are given in numerousStandards.

The Charpy V-notch test for impact resistance is used to measure toughness, thatis, the total energy, elastic and plastic, which can be absorbed by a specimen before

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fracture. The test specimen is a small beam of rectangular cross section with a ‘V’notch at mid-length. The beam is fractured by a blow from a swinging pendulum, andthe amount of energy absorbed is calculated from the loss of height of the pendulumswing after fracture. Details of the test specimen and procedure are given in BS 4360(1990) and BSEN 10045-1 (1990). The Charpy V-notch test is often used to determinethe temperature at which transition from brittle to ductile behaviour occurs.

Structures which are unconventional, and/or method of design which are unusual ornot fully validated by research, should be subject to acceptance tests. Essentially theseconsist of loading the structure to ensure that it has adequate strength to support, forexample, 1 (test dead load) + 1,15 (remainder of dead load) + 1,25 (imposed load).

Where welds are of vital importance, for example, in pressure vessels, they should besubject to non-destructive tests. The defects that can occur in welds are: slag inclusions,porosity, lack of penetration and sidewall fusion, liquation, solidification, hydrogencracking, lamellar tearing and brittle fracture.

A surface crack in a weld may be detected visibly but alternatively a dye can be sprayedonto the joint which seeps into the cracks. After removing any surplus dye the weld isresprayed with a fine chalk suspension and the crack then shows as a coloured line onthe white chalk background. A variant of this technique is to use fluorescent dye anda crack then shows as a bright green line in ultra violet light. A surface crack may alsobe detected if the weld joint area is magnetized and sprayed with iron powder. Thepowder congregates along a crack, which shows as a black line.

Other weld defects cannot be detected on the surface and alternative methods mustbe used. Radiographic methods use an X-ray, or gamma-ray, source on one side ofthe weld and a photographic film on the other. Rays are absorbed by the weld metal,but if there is a hole or crack there is less absorbtion which shows as a dark area onthe film. Not all defects are detected by radiography since the method is sensitive tothe orientation of the flaw, for example, cracks at right angles to the X-ray beam arenot detected. Radiography also requires access to both sides of the joint. The methodis therefore most suitable for in-line butt weld for plates.

An alternative method to detect hidden defects in welds uses ultrasonics. If a weldcontains a flaw then high frequency vibrations are reflected. The presence of a flawcan therefore be indicated by monitoring the reduction of transmission of ultrasonicvibrations, or by monitoring the reflections. The reflection method is extremely usefulfor welds where access is only possible from one side. Further details can be obtainedfrom Gourd (1980).

REFERENCES

Alexander S.J. and Lawson R.M. (1981). Movement design in buildings. Technical Note 107.Construction Industry Research and Information Association Publication.

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BS 4 (2005). Specification for hot rolled sections Pt1. BSI.BSEN 10002-1 (2001). Methods for tensile testing of metals. BSI.BSEN 10045-1 (1990). The Charpy V-notch impact test on metals. BSI.BSEN 10162 (2003). Cold rolled sections. BSI.BS 4360 (1990) Specification for weldable structural steels. BSI.BSEN 10056-1 (1990). Hot rolled structural steel sections. BSI.BSEN 10210-2 (1997). Hot rolled structural steel hollow sections. BSI.BSEN 5950 (2001 and 1998). Structural use of steel work in buildings. BSI.EN 1993-1-1 (2005). General rules and rules for buildings. BSI.EN 1993-1-8 (2005). Design of joints. BSI.Buchanan R.A. (1972). Industrial archeology in Britain. Penguin.CIRIA (1977). Rationalisation of safety and serviceability factors in structural codes. Report 63.

Construction Industry Research and Information Association Publication.Cossons N. (1975). The BP book of industrial archeology. David and Charles.Derry T.K. and Williams T.I. (1960). A short history of technology. Oxford University Press.Gourd L.M. (1980). Principles of welding technology. Edward Arnold.Mann A.P. and Morris L.J. (1981). Lack of fit in steel structures, Technical Report 87. Construction

Industry Research and Information Association Publication.Pannel J.P.M. (1964). An illustrated history of civil engineering. Thames and Hudson.Rolt, L.T.C. (1970). Victorian engineering. Penguin.

Ch02-H5060.tex 11/7/2007 12: 58 page 17

C h a p t e r 2 / Mechanical Properties ofStructural Steel

2.1 VARIATION OF MATERIAL PROPERTIES

All manufactured material properties vary because the molecular structure of thematerial is not uniform and because of inconsistencies in the manufacturing process.The variations that occur in the manufacturing process are dependent on the degreeof control. Variations in material properties must be recognized and incorporated intothe design process.

The material properties that are of most importance for structural design using steelare strength and Young’s modulus. Other properties which are of lesser importanceare hardness, impact resistance and melting point.

If a number of samples are tested for a particular property, for example, strength,and the number of specimens with the same strength (frequency) plotted against thestrength, then the results approximately fit a normal distribution curve as shown inFig. 2.1.

This curve can be expressed mathematically by the equation shown in Fig. 2.1 whichcan be used to define ‘safe’ values for design purposes.

Characteristicstrength

Mean strength

Strength1,64s

x x

y

Fre

quen

cy

5% ofresults

y =1

e−½⎛⎜⎝

⎛⎜⎝

x − xs

2

s√2p

FIGURE 2.1 Variation inmaterial properties

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18 • Chapter 2 / Mechanical Properties of Structural Steel

2.2 CHARACTERISTIC STRENGTH (cl 3.1, EN 1993-1-1 (2005))A strength to be used as a basis for design must be selected from the variation in valuesshown in Fig. 2.1. This strength, when defined, is called the characteristic strength. Ifthe characteristic strength is defined as the mean strength, then clearly from Fig. 2.1,50% of the material is below this value. This is not acceptable. Ideally the characteristicstrength should include 100% of the samples, but this is also impractical because itis a low value and results in heavy and costly structures. A risk is therefore acceptedand it is therefore recognized that 5% of the samples fall below the characteristicstrength.

The characteristic strength is calculated from the equation

fy = fmean − 1,64σ

where for n samples the standard deviation

σ =[

�( fmean − f )2

(n − 1)

]1/2

Nominal values of the characteristic yield strengths and ultimate tensile strengths aregiven in Table 3.1, EN 1993-1-1 (2005) and some examples are given in Table 2.1.

TABLE 2.1 Some nominal values of yield strength for hot rolled steel(Table 3.1, EN 1993-1-1 (2005)).

Standard Nominal thickness of materialEN 10025-2:Grade t ≤ 40 mm 40 < t ≤ 80 mm

fy (MPa) fu (MPa) fy (MPa) fu (MPa)

S235 235 360 215 360S275 275 430 255 410S355 355 510 335 470

TABLE 2.2 Some partial safety factors (cl6.1, EN 1993-1-1 (2005) and Table 2.1,EN 1993-1-8 (2005)).

Situation Symbol Value

Buildings γM0 1,00γM1 1,00γM2 1,25

Joints γM3 1,25γM4 1,00γM5 1,00γM7 1,10

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2.3 DESIGN STRENGTH (cl 6.1, EN 1993-1-1 (2005))The characteristic strength of steel is the value obtained from tests at the rollingmills, but by the time the steel becomes part of a finished structure this strengthmay be reduced (e.g. by corrosion or accidental damage). The strength to be used indesign calculations is therefore the characteristic strength divided by a partial safetyfactor (γM) (Table 2.2).

2.4 OTHER DESIGN VALUES FOR STEEL (cl 3.2.6, EN 1993-1-1(2005))

The elastic modulus for steel (E) is obtained from the relationship between stress andstrain as shown in Fig. 2.2. This is a material property and therefore values from a setof samples vary. However, the variation for steel is very small and the European Codeassumes E = 210 GPa.

The elastic shear modulus (G) is related to Young’s modulus by the expression

E = 2G(1 + ν)

where Poissons ratio ν = 0,3 in the elastic range and is used in calculations involvingplates.

The thermal coefficient of expansion for steel is given as α = 12E-6/K for T ≤ 100◦Cand is used in calculations involving temperature changes.

Fracture

Yield

Str

ess

Fracture

Yield

Str

ess

Strain Strain

Plastic

Elastic

Strainhardening

(a) Low-strength grade S 235 (b) High-strength grade S 450

0,002

FIGURE 2.2 Tensile stress–strain relationships for steel

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Hardness is material property that is occasionally of importance in structural steeldesign. It is measured by the resistance the surface of the steel offers to, the indentationof a hardened steel ball (Brinell test), a square-based diamond pyramid (Vickers test)or a diamond cone (Rockwell test). Higher strength often correlates with greaterhardness but this relationship is not infallible.

Ductility may be described as the ability of a material to change its shape withoutfracture. This is measured by the percentage elongation, that is, 100 × (change inlength)/(original length). Values of 20% can be obtained for mild steel but it is less forhigh-strength steel. A high value is advantageous because it allows the redistributionof stresses at ultimate load and the formation of plastic hinges.

2.5 CORROSION AND DURABILITY OF STEELWORK(cl 4, EN 1993-1-1 (2005))

Durability is a service limit state and the following factors should be considered at thedesign stage:

(a) Environment,(b) Degree of exposure,(c) Shape of the members and details,(d) Quality of workmanship and control,(e) Protective measures,(f) Maintenance.

Methods of protecting steel work are given in BSEN ISO 12944 (1998) and thespecification for weather resistant steel is given is BS 7668 (1994).

Corrosion of steel work reduces the cross-section of members and thus affects safety.Corrosion, which occurs on the surface of steel, is a chemical reaction between iron,water and oxygen, which produces a hydrated iron oxide called rust. Electrons areliberated in the reaction and a small electrical current flows from the corroded areato the uncorroded area.

The elimination of water, oxygen or the electrical current, reduces the rate of corro-sion. In contrast pollutants in the air, for example, sulphur dioxides from industrialatmospheres and salt from marine atmospheres, increase the electrical conductivityof water and accelerate the corrosion reaction.

Steel is particularly susceptible to atmospheric corrosion which is often severe incoastal or industrial environments and the corrosion may reduce the section size dueto pitting or flaking of the surface. Modern rolling techniques and higher-strengthsteels result in less material being used, for example, the web of an ‘I’ section may beonly 6 mm thick. Generally in structural engineering 8 mm is the minimum thickness

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used for exposed steel, and 6 mm for unexposed steel. For sealed hollow sections theselimits are reduced to 4 and 3 mm, respectively.

Corrosion of steel usually takes the form of rust which is a complex oxide of iron. Therust builds up a deposit on the surface and may eventually flake off. The coating of rustdoes not inhibit corrosion, except in special steels, and corrosion progresses beneaththe rust forming conical pits and the thickness of the metal is reduced. The conicalpits can act as stress raisers, that is, centres of high local stress, and in cases wherethere are cyclic reversals of stress, may become the initiating points of fatigue cracksor brittle fracture.

The corrosion resistance of unprotected steel is dependent on its chemical compos-ition, the degree of pollution in the atmosphere, and the frequency of wetting anddrying. Low-strength carbon steels are inexpensive but are particularly susceptible toatmospheric corrosion which is often greatest in industrial or coastal environments.High-strength low-alloy steels (Cr–Si–Cu–P) do not pit as severely as carbon steelsand the rust that forms becomes a protective coating against further deterioration.These steels therefore have several times the corrosion resistance of carbon steels.

The longer steel remains wet the greater the corrosion and therefore the detailing ofsteelwork should include drainage holes, avoid pockets and allow the free flow of airfor rapid drying.

The most common, and cheapest form of protection process is to clean the surface bysand or shot blasting, and then to paint with a lead primer, generally in the workshopprior to delivery on site. Joint contact surfaces need not be protected unless specified.On site the steel is erected and protection is completed with an undercoat and finishingcoat, or coats, of paint.

In the case of surfaces to be welded steel should not be painted, nor metal coated,within a suitable distance of any edges to be welded, if the paint specified or the metalcoating is likely to be harmful to welders or impair the quality of the welds. Weldsand adjacent parent metal should not be painted prior to de-slagging, inspection andapproval.

Encasing steel in concrete provides an alkaline environment and no corrosion will takeplace unless water diffuses through the concrete carrying with it SO2 and CO2 gasesfrom the air in the form of weak acids. The resulting corrosion of the steel and theincrease in pressure spalls the concrete. Parts to be encased in concrete should notbe painted nor oiled, and where friction grip fasteners are used protective treatmentshould not be applied to the faying surfaces.

A more expensive protection is zinc, or aluminium spray coating which is sometimesspecified in corrosive atmospheres. Further improvements are hot dip zinc galvanizing,or the use of stainless steels. These and other forms of protection are described inBSEN ISO 12944 (1998). Recently zinc coated highly stressed steel has been shownto be susceptible to hydrogen cracking.

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2.6 BRITTLE FRACTURE (cl 3.2.3, EN 1993-1-1 (2005))Brittle fracture is critical at the ultimate limit state. Evidence of brittle fracture is asmall crack, which may or may not be visible, and which extends rapidly to producea sudden failure with few signs of plastic deformation. This type of fracture is morelikely to occur in welded structures (Stout et al., 2000).

The essential conditions leading to brittle fracture are:

(a) There must be a tensile stress in the material but it need not be very high, andmay be a residual stress from welding.

(b) There must be a notch, or defect, or hole in the material which produces a stressconcentration.

(c) The temperature of the material must be below the transition temperature(generally below room temperature). At low temperatures crack initiation andpropagation is more likely because of lower ductility.

The mechanism of failure is that the notch, defect or hole raises the local tensilestresses to values as high as three times the average tensile stress. The material whichgenerally fails by a shearing mechanism now tends to fail by a brittle fracture cleav-age mechanism which exhibits considerably less plastic deformation. A drop in thetemperature encourages the cleavage failure. A ductile material which has an exten-sive plastic range is more likely to resist brittle fracture and a test used as a guide toresistance to brittle fracture is the Charpy V-notch impact test (BS 7668 (1994)).

The importance of brittle fracture was shown by the failure of the welded ‘liberty’ cargoships mass produced by the USA during the Second World War. The ships broke apartin harbour and at sea during the cold weather.

Brittle fracture is considered only where tensile stresses exist. The mode of failure ismainly dependent on the following:

(a) Steel strength grade(b) Thickness of material(c) Loading speed(d) Lowest service temperature(e) Material toughness(f) Type of structural element.

No further check for brittle fracture need to be made if the conditions given in EN1993-1-10 (2005) are satisfied for lowest temperature. For further information seeNDAC (1970).

2.7 RESIDUAL STRESSES

Residual stresses are present in steel due to uneven heating and cooling. The stressesare induced in steel during, rolling, welding which constrains the structure to a

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particular geometry, force fitting of individual components, lifting and transporta-tion. These stresses may be relieved by subsequent reheating and slow cooling but theprocess is expensive. The presence of residual stresses adversely affects the bucklingof columns, introduces premature yielding, fatigue resistance and brittle fracture.

Welding raises the local temperature of the steel which expands relative to the sur-rounding metal. When it cools it contracts inducing tensile stresses in the weld andthe immediately adjacent metal. These tensile stresses are balanced by compressivestresses in the metal on either side.

During rolling the whole of the steel section is initially at a uniform temperature, butas the rolling progresses some parts of the cross-section become thinner than othersand consequently cool more quickly. Thus, as in the welded joint, the parts which coollast have a residual tensile stress and the parts which cool first may be in compression.Since the cooling rate also affects the yield strength of the steel, the thinner sectionstend to have a higher yield stress than the thicker sections. A tensile test piece cutfrom the thin web of a Universal Beam will probably have a higher yield stress thanone cut from the thicker flange. The residual stress and yield stress in rolled sectionsare also affected by the cold straightening which is necessary for many rolled sectionsbefore leaving the mills.

Residual stresses are not considered directly in the European Code but are allowedfor in material factors. For further information see Ogle (1982).

2.8 FATIGUE

This is an ultimate limit state. The term fatigue is generally associated with metalsand is the reduction in strength that occurs due to progressive development of existingsmall pits, grooves or cracks when subject to fluctuating loads. The rate of developmentof these cracks depends on the size of the crack and on the magnitude of the stressvariation in the material and also the metallurgical properties. The number of stressvariations, or cycles of stress, that a material will sustain before failure is called fatiguelife and there is a linear experimental relationship between the log of the stress rangeand the log of the number of cycles. Welds are susceptible to a reduction in strengthdue to fatigue because of the presence of small cracks, local stress concentrations andabrupt changes of geometry.

Research into the fatigue strength of welded structures is described by Munse (1984).Other references are BS 5400 (1980), Grundy (1985) and ECSS.

All structures are subject to varying loads but the variation may not be significant.Stress changes due to fluctuations in wind loading need not be considered, but wind-induced oscillations must not be ignored. The variation in stress depends on theratio of dead load to imposed load, or whether the load is cyclic in nature, forexample, where machinery is involved. For bridges and cranes fatigue effects are

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more likely to occur because of the cyclic nature of the loading which causes reversalsof stress.

Generally calculations are only required for:

(a) Lifting appliances or rolling loads,(b) Vibrating machinery,(c) Wind-induced oscillations,(d) Crowd-induced oscillations.

The design stress range spectrum must be determined, but simplified design calcula-tions for loading may be based on equivalent fatigue loading if more accurate data isnot available. The design strength of the steel is then related to the number and rangeof stress cycles. For further information see EN 1993-1-9 (2005).

2.9 STRESS CONCENTRATIONS

Structural elements and connections often have abrupt changes in geometry and alsocontain holes for bolts. These features produce stress concentrations, which are local-ized stresses greater than the average stress in the element, for example, tensile stressesadjacent to a hole are approximately three times the average tensile stress. If the aver-age stress in a component is low then the stress concentration may be ignored, butif high then appropriate methods of structural analysis must be used to cater for thiseffect. The effect of stress concentrations has been shown to be critical in plate webgirders in recent history. Stress concentrations are also associated with fatigue andcan affect brittle fracture. Formulae for stress concentrations are given in Roark andYoung (1975).

2.10 FAILURE CRITERIA FOR STEEL

The structural behaviour of a metal at or close to failure may be described as ductileor brittle. A typical brittle metal is cast iron which exhibits a linear load–displacementrelationship until fracture occurs suddenly with little or no plastic deformation. Incontrast mild steel is a ductile material which also exhibits a linear load–displacementrelationship, but at yield large plastic deformations occur before fracture.

The nominal yield strength is a characteristic strength in the European Code and istherefore an important failure criterion for steel. The tensile yield condition can berelated to various stress situations, for example, tension, compression, shear or variouscombinations of stresses.

There are four generally acceptable theoretical yield criteria as follows:

(1) The maximum stress theory, which states that yield occurs when the maximumprincipal stress reaches the uniaxial tensile stress.

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(2) The maximum strain theory, which states that yield occurs when the maximumprincipal tensile strain reaches the uniaxial tensile strain at yield.

(3) The maximum shear stress theory, which states that yield occurs when themaximum shear stress reaches half of the yield stress in uniaxial tension.

(4) The distortion strain energy theory, or shear strain energy theory, which statesthat yielding occurs when the shear strain energy reaches the shear strain energyin simple tension. For a material subject to principal stresses σ1, σ2 and σ3 it isshown (Timoshenko, 1946) that this occurs when

(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2 = 2fy2 (2.1)

This theory was originally developed by Huber, Von-Mises and Hencky.

Alternatively Eq. (2.1) can be expressed in terms of direct stresses σb, σbc and σbt, andshear stress τ on two mutually perpendicular planes. It can be shown from Mohr’scircle of stress that the principal stresses

σ1 = (σb + σbc)2

−[

(σb − σbc)2

4+ τ2

]1/2

(2.2)

and

σ2 = (σb − σbc)2

+[

(σb − σbc)2

4+ τ2

]1/2

(2.3)

If Eqs (2.2) and (2.3) are inserted in Eq. (2.1) with σ3 = 0 and fy is equal to the designstress fy/γM then

( fy/γM)2 = σ2bc + σ2

b − σbcσb + 3τ2 (2.4)

If σbc is replaced by σbt with a change in sign then

( fy/γM)2 = σ2bc + 3τ2 (2.5)

This equation is a yield criteria which is applicable in some design situations, forexample the design of welds (EN 1993-1-8 (2005)).

REFERENCES

BS 7668 (1994), BSEN 10029 (1991), BSEN 10113-2 (1993), BSEN 10113-3 (1993) and BS 10210-1(1994). Specification for weldable structural steels. BSI.

BS 5400 (1980). Code of practice for fatigue, Pt 10. BSI.BSEN ISO 12944-1 to 8 (1998) and BSEN ISO 14713 (1999). Code of practice for protection of iron

and steel structures. BSI.ECCS (1981). Recommendations for fatigue design of steel structures. European Convention for

Structural Steelwork. Construction Press.EN 1993-1-1 (2005). General rules and rules for buildings. BSI.EN 1993-1-8 (2005). Design of joints. BSI.EN 1993-1-9 (2005). Design of steel structures: Fatigue strength of steel structures. BSI.

Ch02-H5060.tex 11/7/2007 12: 58 page 26


EN 1993-1-10 (2005). Design of steel structures: Selection of steel for fracture toughness and throughthickness properties. BSI.

Grundy, P. (1985). Fatigue limit design for steel structures. Civil Engineering Transactions, Institutionof Civil Engineers, Australia, CE27, No. 1.

Munse, W.H. (1984). Fatigue of welded structures, Welding Research Council.NDAC (1970). Brittle fracture in steel structures, Navy Department Advisory Committee on Structural

Steels (ed G.M. Boyd). Butterworth.Ogle, M.H. (1982). Residual stresses in a steel box-girder bridge. Tech Note 110. Construction Industry

Research and Information Association Publication.Roark, J.R. and Young, W.C. (1975). Formula for Stress and Strain (5th edition). McGraw-Hill.Stout, R.D., Tor, S.S. and Ruzek, J.M. (1951). The effect of fabrication procedures on steels used in

pressure vessels, Welding Journal 30.Timoshenko, S. (1946). Strength of Materials, Pt II. D. Van Nostrand.

Ch03-H5060.tex 11/7/2007 14: 26 page 27

C h a p t e r 3 / Actions

3.1 DESCRIPTION

Actions are a set of forces (loads) applied to a structure, or/and deformations producedby temperature, settlement or earthquakes (EN 1990 (2002)).

Values of actions are obtained by determining characteristic or representative val-ues of loads or forces. Ideally, loads applied to a structure during its working life,should be analysed statistically and a characteristic load is determined. The char-acteristic load might then be defined as the load above which no more than 5% ofthe loads exceed, as shown in Fig. 3.1. However, data is not available and the char-acteristic value of an action is given as a mean value, an upper value or a nominalvalue.

3.2 CLASSIFICATION OF ACTIONS

Actions are classified as:

(1) Permanent(2) Variable(3) Accidental(4) Seismic

Fre

quen

cy

5% of results

Meanload

1,64s

Loads

Characteristicload FIGURE 3.1 Variation in loads

Ch03-H5060.tex 11/7/2007 14: 26 page 28

28 • Chapter 3 / Actions

In addition actions can be classified by:

(1) Variation in time:(a) Permanent actions (G), for example, self-weight and fixed equipment.(b) Variable actions (Q), for example, imposed loads, wind actions or snow loads.

(2) Spacial variation:(a) Fixed actions, for example, structures sensitive to self-weight.(b) Free actions which result in different arrangements of actions, for example,

movable imposed loads, wind actions and snow loads.

3.3 ACTIONS VARYING IN TIME

3.3.1 Permanent Actions (G)

Permanent actions are due to the weight of the structure, that is, walls, permanentpartitions, floors, roofs, finishes and services. The actual weights of the materials (Gk)should be used in the design calculations, but if these are unknown values of densityin kN/m3 may be obtained from EN 1991-1-1 (2002). Also included in this group arewater and soil pressures, prestressing force and the indirect actions such as settlementof supports.

3.3.2 Variable Actions (Q)

(a) Imposed floor loads are variable actions and for various dwellings are given inEN 1991-1-1 (2002). These loads include a small allowance for impact and otherdynamic effects that may occur in normal occupancy. They do not include forcesresulting from the acceleration and braking of vehicles or movement of crowds.The loads are usually given in the form of a distributed load or an alternativeconcentrated load. The one that gives the most severe effect is used in designcalculations.

When designing a floor it is not necessary to consider the concentrated load ifthe floor is capable of distributing the load and for the design of the supportingbeams the distributed load is always used. When it is known that mechanicalstacking of materials is intended, or other abnormal loads are to be applied tothe floor, then actual values of the loads should be used, not those obtained fromEN 1991-1-1 (2002). In multi-storey buildings the probability that all the floorswill simultaneously be required to support the maximum loads is remote andreductions to column loads are therefore allowed.

(b) Snow roof loads are variable actions and are related to access for maintenance.They are specified in EN 1991-1-3 (2002) and, as with floor loads, they areexpressed as a uniformly distributed load on plan, or as an alternative concen-trated load. The magnitude of the loads decrease as the roof slope increases andin special situations, where roof shapes are likely to result in drifting snow, thenloads are increased.

Ch03-H5060.tex 11/7/2007 14: 26 page 29


(c) Wind actions are variable but for convenience they are expressed as static pressuresin EN 1991-1-4 (2002). The pressure at any point on a structure is related to theshape of the building, the basic wind speed, topography and ground roughness.The effects of vibration, such as resonance in tall buildings must be consideredseparately.

(d) Thermal effects need to be considered for chimneys, cooling towers, tanks, hot andcold storage, and services. They are classed as indirect variable actions. Elementsof structures which are restrained or highly redundant introduce stresses whichneed to be determined (EN 1991-1-5 (2003)).

(e) EN 1991-1-2 (2002) covers the actions to be taken into account in the struc-tural design of buildings which are required to give adequate performancein fire.

3.3.3 Accidental Actions (A)

(a) Accidental actions during execution include scaffolding, props and bracing(EN 1991-1-6 (2002)). These may involve consideration of construction loads,instability and collapse prior to the completion of the project.

Erection forces are of great importance in steelwork construction because pre-fabrication is normal practice. Compression members which will be restrainedin a completed structure may buckle during erection when subject to relativelyminor forces. Joints which are rigid when fully bolted may, during erection, actas a pin and induce collapse of the structure. Suspension points for members orparts of structures may have to be specified to avoid damage to components. Itis extremely difficult to anticipate all possible erection forces and the contractoris responsible for erection which should be carried out with due care and atten-tion. Nevertheless a designer should have knowledge of the most likely methodof erection and design accordingly. If necessary temporary stiffening or supportsshould be specified, and/or instructions given.

(b) Accidental actions include impact and explosions which are covered in EN 1991-1-7 (2004). No structure can be expected to resist all actions but it must be designedso that it does not suffer extreme damage from probable actions, for example,vehicle collisions in a multi-storey car park. Local damage from accidental actionsis acceptable.

(c) When designing for earthquakes the inertial forces must be calculated as describedin EN 1998-8 (2004). This is not of major importance in the UK. Actions inducedby cranes and machinery are dealt with in EN 1991-3 (2004).

3.4 DESIGN VALUES OF ACTIONS

Partial safety factors allow for the probability that there will be a variation in the effectof the action, for example, a variable action is more likely to vary than a permanentaction. The values also allow for inaccurate modelling of the actions, uncertainties

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in the assessment of the effects of actions, and uncertainties in the assessment of thelimit state considered.

The design value of an action is obtained by multiplying the characteristic value by apartial safety factor, for example, for a permanent action the design value Gd = γGGk.For a variable action the design value Qd = ψ0Qk or ψ1Qk or ψ2Qk. These representcombination, frequency and quasi-permanent values. The combination value (ψ0Qk)allows for the reduced probability that unfavourable independent actions occur simul-taneously at the ultimate limit state. The frequency value (ψ1Qk) involves accidentalactions and reversible ultimate limit states. The quasi-permanent value (ψ2Qk) alsoinvolves accidental actions and reversible serviceability limit states. Recommendedvalues of ψ0, ψ1 and ψ2 are given in EN 1990 (2002).

3.4.1 Combination of Design Actions

For the ultimate limit state three alternative combinations of actions, modified byappropriate partial safety factors (γ), must be investigated:

(a) Fundamental: a combination of all permanent actions including self-weight (Gk),the prestressing action (P), the dominant variable action (Qk) and combinationvalues of all other variable actions (ψ0Qk).

(b) Accidental: a combination value of the dominant variable actions (ψ0Qk). Thiscombination assumes that accidents (explosions, fire or vehicular impact) of shortduration have a low probability of occurrence.

(c) Seismic: reduces the permanent action partial safety factors (γG) with a reductionfactor (ξ) between 0,85 and 1.

For the serviceability limit state three alternative types of combination of actions mustbe investigated:

(a) The characteristic rare combination occurring in cases when exceeding a limitstate causes permanent local damage or deformation.

(b) The combination which produces large deformations or vibrations which aretemporary.

(c) Quasi-permanent combinations used mainly when long-term effects areimportant.

A combination of actions can be symbolically represented for design purposes, forexample, for one of three conditions at the ultimate limit state:

∑γGGk + γPP +

∑γQψ0Qk

Similar equations can be formed for the other two conditions at ultimate limit stateand for the three conditions at the serviceability limit state (EN 1990 (2002)).

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3.5 ACTIONS WITH SPACIAL VARIATION

3.5.1 Pattern Loading

All possible actions relevant to a structure should be considered in design calculations.The actions should be considered separately and in realistic combinations to determinewhich is most critical for strength and stability of the structure.

For continuous structures, connected by rigid joints or continuous over the supports,vertical actions should be arranged in the most unfavourable but realistic pattern foreach element. Permanent actions need not be varied when considering such patternloading, but should be varied when considering stability against overturning. Wherehorizontal actions are being considered pattern loading of vertical actions need not beconsidered.

For the design of a simply supported beam it is obvious that the critical condition forstrength is when the beam supports the maximum permanent action and maximumvariable action at the ultimate limit state. The size of the beam is then determinedfrom this condition and checked for deflection at the serviceability limit state.

A more complicated structure is a simply supported beam with a cantilever as shownin Fig. 3.2(a).

Assuming that the beam is of uniform section and that the permanent actions areuniformly applied over the full length of the beam, it is necessary to consider variouscombinations of the variable actions as shown in Figs. 3.2(b)–(d). Although partial

A B C

(a)

(b)

(c)

(d)

Imposed design load

Dead design load

Imposed design load

Imposed design load

Dead design load

Dead design loadFIGURE 3.2 Patternloading

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3m

5 m 5 m 5 m

3m

3m

3m

3m

a b

Imposed loads

FIGURE 3.3 Multi-storey frame

loading of spans is possible this is not generally considered except in special cases ofrolling actions (e.g. a train on a bridge span).

For a particular section it is not immediately apparent which combination of actions ismost critical because it depends on the relative span dimensions and magnitude of theactions. Therefore calculations are necessary to determine the condition and sectionfor maximum bending moment and shear force at the ultimate limit state.

Analysis of a multi-storey building is more complicated as shown by Holicky (1996).Where loads on other storeys affect a particular span they may be considered as uni-formly distributed (EN 1991-1-1 (2002)). However, the critical positioning of the loadsmay be different, for example, Fig. 3.3 shows the load positions for verification of thebending resistance at points a and b.

In other situations, for example, when checking the overturning of a structure, thecritical combination of actions may be the minimum permanent action, minimumimposed action and maximum wind action.

3.5.2 Design Envelopes

The effect of pattern loading can be seen by constructing a design envelope. This is agraph showing, at any point on a structural member, the most critical effect that resultsfrom various realistic combinations of actions. Generally the most useful envelopes

Ch03-H5060.tex 11/7/2007 14: 26 page 33


Bending moment (kNm)

2,40 m109(2)

0,21 m

11,7 20,0

38,3

108,3(1)99,0

81,7(2) 91,0

(1) (3)66,4

66,4 (1)(3)

20

87,9

6,9

2,34 m

Shear force (kN)

E F

A B C

Case 1Case 2Case 3Envelope

FIGURE 3.4 Example: Design envelope

are for shear force and bending moment at the ultimate limit state. The formation anduse of a design envelope is demonstrated by the following example.

EXAMPLE 3.1 Example of a design envelope. The beam ABC in Fig. 3.4-carriesthe following characteristic loads:

Dead load Gk = 10 kN/m on both spans;

Imposed loads Qk = 15 kN/m on span AB, 12 kN/m on span BC.

Sketch the design envelope for the bending moment and shear force at the ultimatelimit state. Indicate all the maximum values and positions of zero bending moment(points of contraflexure).

The maximum and minimum design loads on the spans are:

Maximum on AB = γGGk + γQQk = 1,4 × 10 + 1,6 × 15 = 38 kN/m.

Maximum on BC = γGGk + γQQk = 1,4 × 10 + 1,6 × 12 = 33, 2 kN/m.

Minimum on AB or BC = γGGk = 1,0 × 10 = 10 kN/m.

Ch03-H5060.tex 11/7/2007 14: 26 page 34


Consider the following design load cases:

(1) Maximum on AB and BC.(2) Maximum on AB, minimum on BC.(3) Minimum on AB, maximum on BC.

The bending moment and shear force diagrams are shown in Fig. 3.4.

Comments

(a) Only the numerical value of the shear force is required in design, the sign however,may be important in the analysis of the structure.

(b) Positive (sagging) bending moments indicate that the bottom of the beam will bein tension; negative (hogging) moments indicate that the top of the beam will bein tension.

(c) The envelope, shown as a heavy line, indicates the maximum values produced byany of the load cases. Note that on AB the envelope for shear force changes fromcase (2) to case (1) at the point where the numerical values of the shear force areequal.

This process is tedious and an experienced designer knows the critical action combi-nations and the positions of the critical values and avoids some of the work involved.Alternatively the diagrams can be generated from input data using computer graph-ics. In more complicated structures and loading situations, envelopes are useful indetermining where a change of member size could occur and where splices could beinserted. In other situations wind action is a further alternative to combinations ofpermanent and variable actions.

REFERENCES

EN 1990 (2002). Basis of structural design. BSI.EN 1991-1-1 (2002). Densities, self weight and imposed loads. BSI.EN 1991-1-2 (2002). Actions on structures exposed to fire. BSI.EN 1991-1-3 (2002). Snow loads. BSI.EN 1991-1-4 (2002). Wind loads. BSI.EN 1991-1-5 (2003). Thermal actions. BSI.EN 1991-1-6 (2002). Actions during execution. BSI.EN 1991-1-7 (2004). Accidental actions due to impact and explosions. BSI.EN 1998-8 (2004). Design of structures for earthquake resistance. BSI.EN 1991-3(2004). Actions induced by cranes and machinery. BSI.Holicky, M. (1996). Densities, self weight and imposed loads on buildings. Czech Technical University,

Prague.

Ch04-H5060.tex 11/7/2007 17: 5 page 35

C h a p t e r 4 / Laterally RestrainedBeams

4.1 STRUCTURAL CLASSIFICATION OF SECTIONS(CL 5.5, EN 1993-1-1 (2005))

Chapters 4 and 5 are concerned with the design of members which are predominantlyin bending, that is, where axial loads, if any, are small and transverse shear forcesare not excessive. Chapter 4 contains basic theoretical work on section properties andthe design of laterally restrained beams using Class 1 standard sections and plasticmethods of analysis.

Sections of steel beams in common use are shown in Fig. 4.1. The rolled sections shownat (a) are used most often and of these the ‘I’ section is used widely. Some sections areof uniform thickness while others are of different thickness for the web and flange.The rolled sections are generally in stock, are lowest in cost, require less design andconnections are straightforward. Hollow sections are not as efficient in bending butcorrosion resistance is better and aesthetically they may be more acceptable. Coldformed sections are thinner and are therefore more susceptible to corrosion unlessprotected, however they are very economical for use as puriins. Fabricated sectionsare used when a suitable rolled section is not available, but costs are higher anddelivery times are longer. Castellated sections are used for large spans with relativelylow loads and where transverse shear forces are not excessive. Tapered beams areefficient in resisting bending moments but must be checked for shear forces. Compositesteel–concrete sections are used for floors.

The four classes of cross-section of steel ‘I’ beams are described in cl 5.5.2, EN 1993-1-1(2005). To allow for flange buckling sections are reduced to effective sections.

All members subject to bending should be checked for the following at critical sections:

(a) A combination of bending and shear force(b) Deflection(c) Lateral restraint(d) Local buckling(e) Web bearing and buckling.

Ch04-H5060.tex 11/7/2007 17: 5 page 36

36 • Chapter 4 / Laterally Restrained Beams

(a) Hot rolled

(b) Hollow (c) Cold formed, light gauge

Crane rail

Channel weldedto an ‘I’ beam

(d) Fabricated

Cut

Cut

Standard ‘I’ beam

Standard ‘I’ beam

(e) Tapered

Weld

(f) Castellated

(g) Composite

Weld

60°

Reinforcedconcrete

Shear connectors Reinforced concrete

Steel beamencased inconcrete

FIGURE 4.1 Types of steel beams

This chapter is concerned with members which are predominantly subject to bend-ing and where lateral torsional buckling and local buckling of the compression flangeare prevented. It is important to recognize the characteristics of these two formsof buckling shown in Fig. 4.2. Lateral torsional buckling exhibits vertical movement(bending about about the y–y axis), lateral displacement (bending about the z–z axis),

Ch04-H5060.tex 11/7/2007 17: 5 page 37


(b) Local buckling of a flange

Load

z

z

x

x

yy

(a) Lateral torsional buckling of a cantilever

Compression flange

Moment

Moment

Localbuckling

FIGURE 4.2 Buckling of beams

and torsional rotation (rotation about the x–x axis). Local buckling exhibits localdeformation of an outstand, for example, a flange of an ‘I’ beam.

Lateral torsional buckling occurs when the buckling resistance about the z–z axis andthe torsional resistance about the x–x axis are low. The buckling resistance about thez–z axis can be improved by lateral restraints, for example, transverse members whichprevent lateral movement of the compression flange. Local buckling occurs when theflange outstand to thickness ratio (b/tf ) is high and is avoided by choosing Class 1sections (cl 5.5.2, EN 1993-1-1 (2005)).

Where both types of buckling are prevented, as for Class 1 and Class 2 sections, thenthe section can be stressed to the maximum design stresses in bending, i.e. plasticmethods of analysis and design can be used. If Class 3 or Class 4 sections are used theplastic moment capacity is reduced to an elastic moment capacity.

If a steel ‘I’ section is used as a simply supported beam and loaded with a uniformly dis-tributed load then the bending moment distribution varies parabolically. If the sectionis bent about a major axis then the stress distribution at centre span at various stagesof loading is shown in Fig. 4.3(c). In the early stages of loading the stress distribu-tion is elastic, then elastic–plastic and finally fully plastic. The corresponding momentcurvature relationship is shown in Fig. 4.3(b).

The fully plastic stage corresponds to the condition for the tensile stress–strain rela-tionship for the steel shown in Fig. 4.3(a). Theoretically the load cannot be increasedbeyond this plastic condition but strain hardening occurs and this increases the resis-tance. Note that for full plasticity large strains occur, of the order of 20%, whichmakes mild steel ideal, while other steels with less plastic strain behave in a morebrittle fashion.

Although bending is the predominant design criteria checks must be made for themagnitude of the shear stresses. Shear stresses are introduced from vertical shearforces, or torsion moments (cls 6.2.6 and 6.2.7, EN 1993-1-1 (2005)).

Ch04-H5060.tex 11/7/2007 17: 5 page 38


Plastic

Elastic Elastic–plastic

(a) Tensile stress–strain relationship

(c) Change of bending stress distribution as a plastic hinge forms

Strain

Strain hardening

Str

ess

Plastic

Elastic Elastic–plastic

(b) Moment–curvature relationship

(d) Development of a plastic hinge

Neutral axis

Curvature (1/R )

Strain hardening

Mom

ent

sy

FIGURE 4.3 Development of a plastic hinge

For the design of beams calculations are required in the elastic stage of behaviour,for example, stresses and deflections, and also at the fully plastic stage (e.g. collapseload). The calculations involve certain basic section properties.

4.2 ELASTIC SECTION PROPERTIES AND ANALYSIS IN BENDING

4.2.1 Sectional Axes and Sign Conventions

For all standard sections rectangular centroidal axes y–y and z–z are defined parallelto the main faces of the section, as shown in Fig. 4.4. The position of these axes is givenin Section Tables. For angles, and other sections where the rectangular and principalaxes do not coincide, the principal axes are denoted by u–u and v–v. The major axisu–u is conventionally inclined to the y–y axis by an angle α, as shown in Figs 4.4(e)and (f). For equal angles, α = 45◦.

For problems involving simple uniaxial or biaxial bending of symmetrical sections astrict sign convention is not necessary, but for the solution of complex problems it isdesirable. In this chapter the positive conventions of sagging curvature and downwarddeflections are adopted; and the direction of the angle α is anti-clockwise, consistentwith the Section Tables. Fig. 4.5(a) shows the coordinates of a point P in the positive

Ch04-H5060.tex 11/7/2007 17: 5 page 39


zz

z

z

z

z

yy

zz

yyyy

(a) ‘I’ section

(d) Channel (e) Unequal angle (f) Equal angle

(b) Tee (c) Hollow section

y y

z

z

v

v

u

u

ay y

z

z

v

vu

u

y ya � 45°

FIGURE 4.4 Sectional axes

(a) Coordinates and external loads

Fv

MvMz

MuFu

Pu

y

yu

z

z

O

v

α

v

Fy

My

Fz

(b) Stress resultants

Mz

My

VyVy

Vz

Vz

O

O

y

y

z

x

x

z

FIGURE 4.5 Sign conventions

quadrant of a section and the positive directions for the externally applied forces andcouples. The positive directions of the corresponding stress resultants (shear forcesand bending moments) are shown for the horizontal and vertical planes in Fig. 4.5(b).Positive directions relative to the u–u and v–v axes can be inferred. The conventionfor the moments has been chosen so that positive moments give tensile stresses in thepositive quadrant of the section.

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Normally the coordinates of points in a section relative to the rectangular axes areknown, or can easily be obtained. The coordinates relative to the principal axes aregiven by

u = y cos α + z sin α

v = z cos α − y sin α (4.1)

External forces and shear forces transform in exactly the same way, thus

Fu = Fy cos α + Fz sin α

Fv = Fz cos α − Fy sin α (4.2)

However the directions chosen for the moments are consistent with the rules for aright hand set of axes, which gives rise to changes in sign, thus

Mu = My cos α − Mz sin α

Mv = Mz cos α + My sin α (4.3)

4.2.2 Elastic Second Moments of Area

This property is derived from the simple theory of elastic bending (Croxton and MartinVol 1 (1987 and 1989)). In design it is used to calculate stresses and deflections in theelastic stage of behaviour, that is, at service loads. Second moments of area for allstandard sections are given in Section Tables but for fabricated sections they must becalculated. The procedure involves application of the theorems of parallel axes which,for the single element of area A in Fig. 4.6, can be stated as follows:

Iy = Ia + Az2

Iz = Ib + Ay2 (4.4)

Iyz = Iab + Ayz

where

Iy = elemental second moment of area about y–y

Iz = elemental second moment of area about z–z

Iyz = elemental product moment of area about y–y and z–z

a–a and b–b are centroidal axes through the element, parallel to y–y and z–z,respectively.

For the determination of Iyz, which can be either positive or negative, the correct signsmust be allocated to the coordinates y and z. The positive directions are indicated byarrows in Fig. 4.6.

Ch04-H5060.tex 11/7/2007 17: 5 page 41


�y

�z

O

Area A

b

z

b

aac

y

FIGURE 4.6 Parallel axes for an element

When second moments of area about the rectangular axes have been computed, thedirection of the principal axes can be obtained from

tan 2α = 2Iyz/(Iz − Iy) (4.5)

The principal second moments of area are then give by

Iu = Iy cos2 α + Iz sin2 α − Iyz sin 2α

Iv = Iy sin2 α + Iz cos2 α + Iyz sin 2α (4.6)

If Iy is arranged to be greater than Iz, then α will be less than 45◦ and Iu will be themajor principal second moment of area. A negative result for Eq. (4.5) indicates thatα is to be measured clockwise from y–y.

4.2.3 Elastic Section Moduli

These values are derived from the second moments of area by dividing by the distanceto the extreme fibres (i.e. Wel = I/z). Values of section moduli are given in SectionTables. For structural tees two values of Wel are given, referring to the extreme fibresin the table and the stalk.

4.2.4 Elastic Bending of Symmetrical Sections

When either of the rectangular axes is an axis of symmetry the normal bending stressat any point in the section is given by

σ = MxzIy

+ MzyIz

(4.7)

If the directions of the bending moments and the coordinates are in accordance withthe sign convention of Fig. 4.5, a positive result indicates that the stress is tensile.

Ch04-H5060.tex 11/7/2007 17: 5 page 42


For simple bending about the y–y axis, which is the most common event, the stress inthe extreme fibres

σmax = Myzmax

Iy= My

Wely(4.8)

Similarly for bending about the z–z axis

σmax = Myzmax

Iy= My

Wely(4.9)

4.2.5 Elastic Bending of Unsymmetrical Sections

When a section is subject to bending about an axis which is not a principal axis theeffect is the same as if the section were subject to the components of the bendingmoment acting about the principal axis. In other words the bending is biaxial.

For standard rolled angles the principal second moments of area and the directionsof the principal axes are given in Section Tables. Transforming bending moments andcoordinates to the principal axes by means of Eqs (4.3) and (4.1). Bending stress

σ = Muv

Iu+ Mvu

Iv(4.10)

This is the same as Eq. (4.7) but with all the terms treated to the principal axes. If thesign convention of Fig. 4.5 is observed, a positive result indicates tension.

In other cases the additional calculations required for the solution of problems byprincipal axes can be avoided by the use of ‘effective bending moments’. These aremodified bending moments which can be considered to act about the rectangular axesof the section. The bending stress is then given by an expression having exactly thesame from as Eq. (4.7)

σ = MeyzIy

+ MezyIz

(4.11)

where Mey and Mez are effective bending moments about y–y and z–z axes, respectivelyand are given by

Mey = My − MzIyz/Iz

1 − Iyz2/(IyIz)

Mez = Mz − MyIyz/Iy

1 − Iyz2/(IyIz)(4.12)

These expressions are derived from the application of conventional elastic bendingtheory to curvature in both the yx and zx planes. One such derivation is given byMegson (1980).

Ch04-H5060.tex 11/7/2007 17: 5 page 43


By successive differentiation with respect to x, the longitudinal dimension, sim-ilar expressions for the effective shear force and effective load intensity can beobtained, thus

Vey = Vy − VzIyz/Iy

1 − Iyz2/(IyIz)

Vez = Vz − VyIyz/Iz

1 − Iyz2/(IyIz)(4.13)

and

fey = ( fy − fzIyz/Iy)1 − Iyz2/(IyIz)

fez = ( fz − fyIyz/Iz)1 − Iyz2/(IyIz)

(4.14)

It should be noted that the quantities Iy and Iz are interchanged in Eqs (4.13) and(4.14). This is because the expressions for the shear force and the load intensity in they direction are obtained by successive differentiation of bending moments along thez–z axis and vice versa.

All bending moment problems with unsymmetrical sections can be solved simply byreplacing ordinary loads, shears, and bending moments by their effective counterparts.Note however that these effective counterparts have values related to both the y–y andz–z axes, even if the section is only loaded in the direction of one of the rectangularaxes.

EXAMPLE 4.1 Principal axes for an unequal angle section. Find the directions ofthe principal axes and the values of the principal second moments of area for the anglesection in Fig. 4.7(a).

For the calculation of section properties the work is simplified considerably, withinsignificant loss of accuracy, by using the dimensions of the section profile, that is,the shape formed by the centre line of the elements, as shown in Fig. 4.7(b).

The position of the centroid O is found by taking moments of the area about the centrelines of each leg in turn

Areas mm2

A′B′ 140 × 20 = 2800A′C′ 290 × 20 = 5800

Total = 8600

Ch04-H5060.tex 11/7/2007 17: 5 page 44


20

300

20

150 140

B� A�

c �z

c� y

c�

cz

c y

Oa

y

u

O

290

z v

(b) Section profile(a) Actual section

FIGURE 4.7 Example: principal axes for an unequal angle

Taking moments about A′B′

8600c′y = 5800 × 290

2hence c′

y = 97,8 mm

Taking moments about A′C′

8600c′z = 2800 × 140

2hence c′

z = 22,8 mm

Hence for the full section (Fig. 4.7(a))

cy = 107,8 mm and cz = 32,8 mm

Coordinates of the centroids of the legs AB and AC are therefore given by

Leg A′B′ y = 1402

− c′z = 47,2 mm

z = −c′y = −97,8 mm

Leg A′C′ y = −c′z = −22,8 mm

z = 2902

− c′y = 47,2 mm

The second moments of area about the rectangular axes are obtained in the usual waybe applying the parallel axes formula to each leg.

Iy(leg) = bh3

12+ A(leg)z2

where b and h are dimensions of the leg parallel to the y–y and z–z axes, respectively.

Ch04-H5060.tex 11/7/2007 17: 5 page 45


Leg A′B′ 140 × 203

12+ 2800 × (−97,8)2 = 26,87E6 mm4

Leg A′C′ 20 × 2903

12+ 5800 × 47,22 = 53,57E6 mm4

Iy = 80,44E6 mm4

Iz(leg) = hb3

12+ A(leg)y2

Leg A′B′ 20 × 1403

12+ 2800 × 47,22 = 10,81E6 mm4

Leg A′C′ 290 × 203

12+ 5800 × (−22,8)2 = 3,21E6 mm4

Iz = 14,02E6 mm4

The product moment of area Iyz is obtained by applying the parallel axis formula toeach leg

Iyz(leg) = Iab + A(leg)yz

For each leg the term Iab is equal to zero, because the parallel axes through the centroidof the leg are principal axes.

Leg A′B′ 2800 × 42,7 × (−97,8) = −12,93E6 mm4

Leg A′C′ 5800 × (−22,8) × 47,2 = −6,24E6 mm4

Iyz = −19,17E6 mm4

Direction of the principal axes from Eq. (4.5)

tan 2α = 2Iyz

Iz − Iy

2α = arctan[

2 × (−19, 17)14,02 − 80,44

]hence α = 15◦

Principal second moments of area from Eq. (4.6)

Iu = Iy cos2 α + Iz sin2 α − Iyz sin 2α

Iv = Iy sin2 α + Iz cos2 α + Iyz sin 2α

Substituting values

Iu = 85,58E6 mm4 and Iv = 8,88E6 mm4.

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50 kN m

100 kN m

bf

c y

t f

tw

z

z

P

R

Q

hy y

FIGURE 4.8 Example: structural tee in biaxialbending

As a check on the transformation

Iy + Iz = Iu + Iv which is correct.

EXAMPLE 4.2 Structural tee in biaxial bending. Calculate the maximum extremefibre stresses in a standard 292 × 419 × 113 kg structural tee cut from a UniversalBeam. The tee is loaded by two moments as shown in Fig. 4.8.

From Section Tables

bf = 293,8 mm, h = 425,5 mm, tw = 16,1 mm, tf = 26,8 mm, Cy = 108 mm,Iy = 246,6E6 mm4, Iz = 56,76E6 mm4, Wely(flange) = 2,277E6 mm3,Wely(toe) = 0,7776E6 mm3, Welz = 0,3865E6 mm3.

By inspection the maximum compressive stress occurs at a point P because the stressesfrom both moments are compressive.

σ = My

Wely(flange)+ Mz

Welz

= 100E62,277E6

+ 50E60,3865E6

= 173 MPa

The maximum tensile stress can occur at point Q or point R, depending on the relativemagnitude of the bending moments. It is necessary to check both points.

Using the sign convention of Fig. 4.5 both bending moments are positive and thecoordinates of the points are given by

Point Q, y = tw2

= 16,12

= 8,05 mm, z = h − cy = 425,5 − 108 = 317,5 mm.

Point R, y = bf

2= 293,8

2= 146,9 mm, z = tf − cy = 26,8 − 108 = −81,2 mm.

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From Eq. (4.7) the stresses

σ = MyzIy

+ MzyIz

σQ = 100 × 317,5246,6

+ 50 × 8,0556,76

= 135,8 MPa

σR = 100 × (−81,2)246,6

+ 50 × 146,956,76

= 96,5 MPa

Summarizing, the maximum stresses are:

At Q, σQ = +135,8 MPa (tension)

At P, σP = −173 MPa (compression)

EXAMPLE 4.3 Bending stresses in an unequal angle section. Calculate the bendingstresses in the angle section shown in Fig. 4.9 where

Iy = 80,44E6 Iz = 14,02E6 Iyz = −19,17E6 mm4

Effective moments from Eq. (4.12)

Mey = (My − MzIyz/Iz)(1 − Iyz2/IyIz)

= 74,92 kNm

Mez = (Mz − MyIyz/Iy)(1 − Iyz2/IyIz)

= 32,86 kNm

Maximum compressive stress at A

σA = MeyzA

Iy+ MezyA

Iz

= 74,92E6 × (−107,8)80,44E6

+ 32,86E6 × (−32,8)14,02E6

= −177,3 MPa (compression)

Check for maximum tensile stress at B, and at C

σB = MeyzB

Iy+ MezyB

Iz

= 74,92E6 × (−87,8)80,44E6

+ 32,86E6 × 117,214,02E6

= +192,9 MPa (tension)

σC = MeyzC

Iy+ MezyC

Iz

= 74,92E6 × 192,280,44E6

+ 32,86E6 × (−12,8)14,02E6

= +149,0 MPa (tension)

Ch04-H5060.tex 11/7/2007 17: 5 page 48


150

117,2

30 kN m

15 kN m

32,8

87,8

12,8

z(a)

C

y

192,

2

300

107,

8

A

B

183,6

155

78,4

Neutral axis

C�125,6

�144,5

B�A�

(c)

Mz

Mz

My

My

y

z

x

(b) Positive bending moments

FIGURE 4.9 Example: bending stresses in an unequal angle

Summarizing, the maximum stresses are:

At B, σB = +192,9 MPa (tension)

At A, σA = −177,3 MPa (compression)

Note that although the position of the centroid and the values of the second momentof area can be calculated without significant error from the profile dimensions of thesection, the same is not true of the stresses.

EXAMPLE 4.4 Bending about principal axes of an angle section. Recalculate thestresses at points A, B and C in the previous example considering bending about theprincipal axes.

Ch04-H5060.tex 11/7/2007 17: 5 page 49


The coordinates of the points are transformed in accordance with Eq. (4.1),

u = y cos α + z sin α and v = z cos α − y sin α.

sin α = 0, 2588 and cos α = 0, 9659,

Point y/z axes u/v axes

y z u v

A −32,8 −107,8 −59,6 −95,6B 117,2 −87,8 90,5 −115,1C −12,8 192,2 37,5 189,0

Bending moments transform in accordance with Eq. (4.3)

Mu = My cos α − Mz sin α = 25,10 kNm

Mv = Mz cos α + My sin α = 22,25 kNm

Bending stresses from Eq. (4.10)

σ = Muv

Iu+ Mvu

Iv, σA = −177,3 σB = 192,9 σC = 149,0 MPa

which are the same as obtained previously.

4.2.6 Elastic Analysis of Beams

The elastic analysis of simply supported beams with examples of shear force and bend-ing moment diagrams and deflection calculations are given in many introductory bookson structural analysis (e.g. Croxton and Martin Vol. 1 (1987 and 1989)). The analysis ofcontinuous beams is more complicated but there is a choice of methods such as area–moment, moment distribution, slope deflection, and matrix methods. These methodsare also covered by Croxton and Martin Vol. 2 (1987 and 1989). with a computerprogram for analysis using matrix methods.

4.2.7 Elastic Deflections of Beams (cl 7.2.1, EN 1993-1-1 (2005))

The deflections under serviceability loads of a building or part should not impairthe strength or efficiency of the structure or its components or cause damage to thefinishes. When checking for deflections the most adverse realistic combination andarrangement of serviceability loads should be assumed, and the structure may assumedto be elastic.

The theory and the methods of calculating deflections for static and hyperstatic struc-tures are given in Croxton and Martin Vols. 1 and 2 (1987 and 1989). For simplebeams standard cases can be superimposed and some useful cases are shown inFigs 4.10 and 4.11.

Ch04-H5060.tex 11/7/2007 17: 5 page 50


Q

(a)

a2(3 � a)QL3/(6EI )

a3(4 � a)qL4/(24EI )

a3(5 � a)qL4/(120EI )

(b)

(c)

aL

aL

aL

L

q

q

Maximum deflection (at free end)

FIGURE 4.10 Deflections ofcantilevers

For simply supported beams the central deflection rather than the maximum is given,so that deflections from individual cases can be added. For most loading cases thecentral deflection only differs by a small percentage from the maximum. In case (a) ofFig. 4.11, for example, the difference is always within 2,5%. A notable exception is thecase of equal end moments acting in the same direction, when the central deflection iszero. However in such a case the deflection at other points along the beam are likely tobe small. A more accurate analysis can be formed if it is suspected that the deflectionis likely to exceed the limit. Recommendations for limiting values for deflections aregiven in cl 7.2.1, EN 1993-1-1 (2005).

EXAMPLE 4.5 Deflections for a hyperstatic structure. The size of the members forthe symmetrical structure shown in Fig. 4.12 have been determined and the structurerequires to be checked for deflections in the elastic range of behaviour. The imposedvariable characteristic loads are shown in Fig. 4.12 and the second moment of area isIy = 127,56E6 mm4. The bending moments (positive clockwise) at the joints are givenin the following table.

Joint Span Moment (kNm)

B AB +60B BC −60C BC −6C CG +72C CD −66G CG +36

Ch04-H5060.tex 11/7/2007 17: 5 page 51


Q

(a)

(b)

q

(c)

q(d)

(e)

q

aL

End 1 End 2

aL

L

Central deflection, androtation at supports

For a � 0.5d � (3a � 4a3)QL3/(48EI )

u1 � (2a � 3a2 � a3)QL2/(6EI )

u2 � (a3 � a)QL2/(6EI )

For a � 0.5d � (3a2 � 2a4)qL4/(96EI )

u1 � (a4 � 4a3 � 4a2)qL3/(24EI )u2 � (a4 � 2a2)qL3/(24EI )

d � 5qL4/(384EI )u1 � qL3/(24EI )u2 � �u1

d � qL4/(120EI )u1 � 5qL3/(192EI )u2 � �u1

d � (M1�M2)L2/(16EI )u1 � (2M1�M2)L /(6EI )u2 � (2M2�M1)L /(6EI )

M1 M2

FIGURE 4.11Displacements of simplysupported beams

3 3184 6

8

18 4

A

Dimensions in m; loads in kN.

B FC

G H

D E30 36 30

24 24

FIGURE 4.12 Example: deflections of a symmetrical continuous structure

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For all beams

EIy = 210E3 × 127,56E6 = 26,79E12 N mm2

Span CD

1. Uniform load (Fig. 4.11(b))

∂ul = 5qL4

384EIy= 5QL3

384EIy

= 5 × 36E3 × 12E33

384 × 26,79E12= 30,2 mm

2. Concentrated loads (Fig. 4.11(a))

∂cl = 2[3a − 4a3]QL3

48EIy

= 2[3 × (3/12) − 4 × (3/12)3] × 24E3 × 12E33

48 × 26,79E12= 44,3 mm.

3. End moments (Fig. 4.11(e))

∂M = L2 (M1 − M2)16EIy

= 12E32 × (−66 − 66)E616 × 26,79E12

= −44,3 mm (upwards)

Total deflection = ∂ul + ∂cl + ∂M = 30,2 + 44,3 + (−44,3) = 30,2 mm

A limit for beams with plaster finish = L/350 = 12E3/350 = 34,3 mm

Span BC

End moments (Fig. 4.11(e))

∂3 = L2 (M1 − M2)16EIy

= 18E32 × (−60 + 6)E616 × 26,79E12

= −40,8 mm (upwards)

An accurate analysis gives −42,4 mm at 7,35 m from B,

A limit for beams with plaster finish = L/350 = 18E3/350 = 51,43 mm

Cantilever span AB

For this span the deflection is due to the flexure of the cantilever, assuming the beamis horizontal at B, plus the effect of the anti-clockwise rotation of the beam at B,that is, −θ1 for span BC.

Ch04-H5060.tex 11/7/2007 17: 5 page 53


End moments for span BC (Fig. 4.11(e))

θ1 = L × (2M1 − M2)6EIy

= 18E3 × (−2 × 60 + 6)E66 × 26,79E12

= −0,01277 rad

1. Deflection at A due to rotation

∂R = −Lθ1 = −4E3 × (−0,01277) = 51,07 mm

2. Deflection due to load (Fig. 4.10(b))

∂ul = a3(4 − a)qL4

24EI= QL3

8EIy

= 30E3 × 4E33

8 × 26,79E12= 8,96 mm

Total deflection = ∂R + ∂ul = 51,1 + 9,0 = 60,1 mm

A limit for a cantilever beam with plaster finish

2L350

= 2 × 4E3350

= 22,9 mm

If the deflection of the cantilever exceeds the limit and stiffening is required thenincrease the size of the section, or add flange plates. It is also necessary to stiffenspans AB and BC because the deflection is dependent on both.

4.2.8 Span/Depth Ratios for Simply Supported Beams

An initial estimate for the depth of a simply supported ‘I’ beam carrying a uniformlydistributed load can be obtained by using the deflection limit. If σmax is the maximumelastic bending stress at service load then from elastic bending theory

σmax = Mz

Iy= QL

8× h/2

Iy

rearranging

Q = 16σmaxIy

Lh(i)

Assuming a deflection limit for beams with plaster finish of

L350

= 5QL3

384EIy(ii)

Eliminating Q by combining Eqs (i) and (ii) and putting E = 210 GPa, the span depthratio

Lh

= 2880σmax

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If a beam is laterally restrained so that lateral torsional buckling does not occur, then ifthe maximum bending stress is approximately fy/1,5, the span/depth ratios for differentgrades of steel and different limits are:

TABLE 4.1 Span/depth ratios for beams.

Span/depth ratioCharacteristic

Grade stress (fy) L/350 L/250

S235 235 18,4 25,7S275 275 15,7 22,0S355 355 12,2 17,0

Note that since Youngs modulus (E) is a constant for all grades of steel, the stiffnessof a beam does not increase with a higher grade of steel. If a design is governed bydeflection then there is no advantage in using a higher grade of steel.

4.3 ELASTIC SHEAR STRESSES

4.3.1 Elastic Shear Stress Distribution for a SymmetricalSection

When a beam is bent elastically by a system of transverse loads, plane sections nolonger remain plane after bending, but are warped by shear strains. In most cases theeffect is small and the errors introduced in the use of conventional bending theoryare negligible. Important exceptions are discussed briefly in Section 4.3.2. Formulaefor the calculation of shear stresses in an elastic beam are derived by considering thevariation in bending stresses along a short length of beam.

Consider the very short length of beam in Fig. 4.13(a). At a point S in the web theshear stresses on the vertical and the horizontal section are complementary and aregiven by the established formula (Eq. (6.20), EN 1993-1-1 (2005))

vs = VAzc

It(4.15)

where

V = the vertical shear force on the section

A = the hatched area, that is, the part of the section between point S and the extremefibres

zc = the distance from the centroid of the area A to the neutral axis

I = the second moment of area of the whole section about the neutral axis

t = the thickness of the section at the point S.

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b

P

NA

z c

S

tw

vs

vs

tf

Area A

(a) Shear flow

(b) Stress distribution

P

bf

tf

vp

vs

tw

S

h

FIGURE 4.13 Shearstresses in an ‘I’ beam

The formula cannot be used to obtain the vertical shear stress in the outstanding partsof the flange. However as this must be equal to zero at the top and bottom faces, itmust be very small. In fact the resistance of the section to vertical shear is providedalmost entirely by the web.

The resultant of the longitudinal shear stress in the web is in equilibrium with thechange in the normal tensile force on the area A due to the variation in bendingmoment along the beam. Similar longitudinal stresses exist in the flanges and give riseto horizontal complementary stresses in the directions shown. For example, at pointP in the top flange.

A = bf tf , tf = t, zc = (h − tf )2

and Eq. (4.15) becomes

vp = Vb(h − tf )2I

(4.16)

This expression is linear with respect to the variable b, and vp has a maximum valueat the centre of the flange where b = bf /2, that is

vp(max) = Vbf (h − tf )4I

(4.17)

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The complete distribution of shear stress on the cross-section is shown in Fig. 4.13(b).

Equation (4.15) can be expressed in terms of the shear flow, which is the product ofthe shear stress and the thickness of the section, thus

v = tvs = VAzc

I(4.18)

In the longitudinal sense the shear flow is equal to the shear force per unit length ofbeam, and is a convenient quantity for the calculation of the shear force to be resistedby bolts or welds in a fabricated section.

4.3.2 Elastic Shear Stresses in Thin Walled Open Sections

The shear stress distribution for a rectangular cross-section subject to a transverseshear force is shown in Fig. 4.14(a) and can be calculated using Eq. (4.15). The shearstress distribution for the open cross-section is different as shown in Fig. 4.14(b).the distribution for an ‘I’ section is shown in Fig. 4.14(c). Steel sections are usuallycomposed of relatively thin elements, for which the analysis can be simplified by:

(a) referring all dimensions to the profile of the section;(b) assuming that the shear stress does not vary across the thickness;(c) ignoring any shear stresses acting at right angles to the section profile. As these

are equal to zero at each outside surface they must always be very small in a thinwalled section.

If it is further assumed that the load is applied in such a way that no twisting of thebeam occurs, the shear flow at a point S on the profile of the section is given by

vs = vo −(

Vez

Iy

)∫ s

otz ds −

(Vey

Iz

)∫ s

oty ds (4.19)

here Vey and Vez are the effective shear forces obtained from Eq. (4.13) or by applyingthe effective loads obtained from Eq. (4.14). The variable s is the distance aroundthe profile to the point of interest, starting from any point at which the shear flow

(a) (b) (c)

FIGURE 4.14Distribution of shearstresses

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vo is known. At any open end, such as the end of a flange, the value of vo is zero.The direction of s can be chosen arbitrarily and, provided that the sign convention ofFig. 4.5 is adopted, a positive sign for vs indicates that the shear flow is in the directionchosen for s.

4.3.3 Elastic Shear Stresses in Thin Walled Closed Sections

The shear stress and shear flow in a symmetrical closed section can be obtained directlyfrom Eqs (4.15) and (4.18), respectively. For an unsymmetrical section Eq. (4.19) canbe used, but analysis is complicated by the fact that that vo is not known at any point.The problem can be solved by first cutting the section at some point and finding theposition of the shear centre in the resulting open section. The shear flow in the closedcross-section then results from the combined action of the applied shear loads trans-ferred to the shear centre of the cut (open) section, and the torque on the closedsection due to the transference of loads. Examples of shear stress distribution aregiven in Fig. 4.14. A similar approach is used to find the position of the shear centreof the closed section.

4.3.4 Elastic Shear Lag (cl 6.2.2.3, EN 1993-1-1 (2005))

The simple theory of bending is based on the assumption that plane sections remainplane after bending. In reality shear strains cause the section to warp. The effect inthe flanges is to modify the bending stresses obtained by the simple theory, producinghigher stresses near the junction of a web and lower stresses at points remote fromit as shown in Fig. 4.15. This effect is described as ‘shear lag’. The discrepancies pro-duced by shear lag are minimal in rolled sections, which have relatively narrow andthick flanges. However in plate girders, or box sections, having wide thin flanges theeffects can be significant when they are subjected to high shear forces, especially inthe vicinity of concentrated loads where the sudden change in shear force produceshighly incompatible warping distortions. Shear lag effects can be allowed for by usingeffective widths (cl 3, EN 1993-1-5 (2003)).

Bending stressnon-uniform

Warpingdisplacements

Shear force

FIGURE 4.15 Shearlag effects for an ‘I’section

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(a) (b) (c) (d)

Shear centre CentroidFIGURE 4.16 Position of shearcentre

Vz

Vy

C

D

NB

A

y′

Shear centreCentroid

FIGURE 4.17 Shear centre–unsymmetrical section

4.3.5 Elastic Shear Centre

Equation (4.19) is only valid if no twisting of the beam occurs at the section considered.Torsion in a section can be generated by a transverse load if the resultants of the shearstresses in the elements of the section produce a torque. To counteract this the lineof action of the applied load must pass through the shear centre. In a symmetricalsection the shear centre lies on an axis of symmetry, and loads applied along such anaxis do not cause twisting. In some sections the position of the shear centre can beinferred from the direction of the shear flow (Fig. 4.16). In (a) the shear center liesat the intersection of the two axes of symmetry and is coincident with the centroid; in(b) and (c) it lies at the intersection of lines of shear flow; in (d), if the flanges are ofthe same size, the shear stresses in them set up opposing torques about the centroidwhich is therefore the shear centre.

For the general case of an unsymmetrical thin walled open section subject to biaxialbending, with shear forces Vy and Vz, the position of the shear centre can be found bydetermination of the shear flow from Eqs (4.18) or (4.19), applying Vy and Vz in turn,assuming that they pass through the shear centre. Consider, for example, the sectionprofile in Fig. 4.17. If point B is chosen as the fulcrum it is only necessary to find theresultant shear forces in the leg CD due to Vy and Vz in turn. These forces producetorques equal to Vyz′ and Vzy′, respectively. By taking moments about B the values of

Ch04-H5060.tex 11/7/2007 17: 5 page 59


z′ and y′ can be obtained. There is no need to calculate the shear stresses in AB or BCbecause their lines of action pass through the point B, and generate no moment. Theresultant shear forces in CD are obtained by integrating the shear stresses obtainedby Eq. (4.19) along the leg.

The above process is tedious since, for each value of Vy and Vz, the correspondingeffective shear forces Vey and Vez must be calculated and applied. If there is an axis ofsymmetry Eq. (4.18) can be used and the analysis is simplified.

EXAMPLE 4.6 Distribution of shear stresses for an angle section. Calculate theshear stresses in the simply supported angle section shown in Fig. 4.18(a). Iy = 80,44E6,Iz = 14,02E6, Iyz = −19,17E6 mm4.

To the left of mid-span the shear forces are

Vy = 5 and Vz = 10 kN

20 kN

5

10 kN

10

10

5

6 m span

(a)

(c)

(b)

(d)

Vy

Vy

Vz

Vz

117,222,8

97,8

192,

2140

290

s1

s2

z

z

y

y

78,4

2,4 (max)

2,8(max)

155,

2

0,920,3

FIGURE 4.18Example: distributionof shear stresses for anunsymmetrical section

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Effective shear forces from Eq. (4.13)

Vey = (Vy − VzIyz/Iy)1 − Iyz2/IyIz

= 10,95 kN

Vez = (Vz − VyIyz/Iz)1 − Iyz2/IyIz

= 24,97 kN

Shear flow from Eq. (4.19)

vs = vo −(

Vez

Iy

)∫ s

otz ds −

(Vey

Iz

)∫ s

oty ds (i)

For the horizontal leg starting from the left hand end

vo = 0, s = s1, y = (117,2 − s1) and z = −97,8 mm

Substituting these values in (i) and integrating

vs1 = 0,00781s21 − 1,224s1 (ii)

This equation shows that vs1 = 0 only when s1 = 0. Differentiating with respect tos1 and equating to zero gives a turning point at s1 = 78,4 mm. Hence from (ii)vs1(max) = −47,96 N/mm, and s1 = 140 mm, vs1 = −18,28 N/mm. The negative signsindicate that the shear flow is in the opposite direction to s1.

For the vertical leg

s = s2, y = −22,8 mm, z = (s2 − 97,8) mm, vo = vA = −18,28 N/mm.

Substitution of these values into (i) gives

vs2 = −18,28 + 0,9633s2 − 0,003104s22 (iii)

Solving (iii) for s2 shows that when vs2 = 0, s2 = 20,3 mm. There is also a turning pointat s2 = 155,2 mm. Hence from (iii) vs2(max) = 56,46 N/mm. The positive sign indicatesthat the shear flow is in the direction of s2. As a check, putting s2 = 290 mm givesvs2 = 0, which is correct.

The shear stresses in MPa are obtained by dividing the shear flows by the thickness,that is, 20 mm, and are plotted for the whole section in Fig. 4.18(d).

This example demonstrates the method of analysis but the shear stresses are very lowand do not justify such a detailed treatment. As a rough guide as to whether detailedanalysis is required the shear forces are divided by the area of the appropriate leg.

If the shear stress is only required at particular points in the section Eq. (4.15) can beused with the effective shear forces, taking each axis in turn and superimposing theresults. Integration is avoided but the directions of the stresses have to be found byinspection. It can be seen from the position of the neutral axis in Example 4.3 thatthe maximum shear stresses occur where the neutral axis intersects the profile of the

Ch04-H5060.tex 11/7/2007 17: 5 page 61


y � b

Ayy d

t

v z

Shear centre FIGURE 4.19 Example: shear centre of a channel

section, as in a symmetrical section. If these points have previously been found, thenthe maximum shear stress can be calculated directly as above, using Eq. (4.15).

EXAMPLE 4.7 Shear centre for a channel section. Find the position for the shearcentre for the channel shown in Fig. 4.19 which has a uniform thickness.

As the shear centre lies on the axis of symmetry y–y, there is no need to consider Vy.If point A at the intersection of axis y–y with the centre line of the web is taken asthe fulcrum, then only the shear force in the flanges need to be considered since theresultant shear force in the web produces no moment about A. Equation (4.18) is

vs = VAzc

I

The distribution of shear flow in the flanges in linear with zero at the ends andmaximum at the web centre line. Hence, for maximum shear flow

A = bt, V = Vz, zc = d2

and I = Iy, which gives

vs(max) = Vzbtd2Iy

The resultant shear force is equal to half the maximum shear flow multiplied by theflange width, that is

Force = Vzb2td4Iy

The torque about point A from both flanges is equivalent to the torque produced bythe applied shear force when it passes through the shear centre, thus

Vzy′ = Vzb2td2

4Iy

from which

y′ = b2d2t4Iy

(4.20)

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4.4 ELASTIC TORSIONAL SHEAR STRESSES (CL 6.2.7,EN 1993-1-1 (2005))

4.4.1 Elastic Torsion

Generally torsion is not a major problem in the design of beams except for specialcases. Torsional shear stresses arise from a variety of causes including, beams crankedor curved in plan, and loads whose line of action does not pass through the shearcentre of the section. It is important to realize that in torsional problems the lever armfor the torque is measured from the shear centre, not the centroid. In particular, thedistributed loads on unsymmetrical sections, such as angles or channels, usually actthrough the centroid of the section, as in the case of self-weight, although generallyself-weight does not produce large torsional stresses. However, if other loads are offsetfrom the shear centre the effect can be considerable, for example, for a 381 × 102 chan-nel section the lever arm for loads acting through the centroid is approximately 50 mm.

The total torsional moment at any section is the sum of the elastic torsion (St. Venant)plus the internal warping torsion. The warping torsion consists of a bi-moment pluswarping.

4.4.2 Uniform and Non-uniform Elastic Torsion

In general the cross-sections of members subject to torsion do not remain plane, buttend to warp. Warping is the change in geometric shape of the member so that cross-sections do not remain plane as shown in Fig. 4.20. The degree of warping that takesplace depends on the shape of the section, and is most pronounced in thin walledchannels. In some sections, such as angles and tees, solid and hollow circular sectionsand square box sections of uniform thickness, warping is virtually non existent, whilein others, such as closed box sections of general shape, its effect is small. In ‘I’ sectionsmost of the warping takes place in the flanges and its effect on the web is very smalland can be ignored. If torque is applied only at the ends of the member and warping isnot restrained, the flanges remain virtually straight and maintain their original shapeas shown in Fig. 4.20(a). The result is that the sectional planes of the flanges rotatein opposite directions, producing warping displacements which are constant along thewhole length of the member. Under these circumstances the member is said to be inthe state of uniform, or St. Venant, torsion.

If warping is prevented, for example, by rigid supports at the ends of the member,the torsional stiffness is increased and longitudinal tensile and compressive stressesare induced. In practice warping is not so obvious but may arise from the action ofstructural connections, or from the incompatibility of warping displacements that occurwhen the torque is not uniform along the length of the member. Warping restraintincreases the torsional stiffness of a member, and at any point along its length theapplied torque is resisted by the two components, one due to the St. Venant torsion,

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Mb

Mb

T

Stress

T

Stress

h

Mx

tfH

H

(a)

(c)

(d) (e)

(b)

FIGURE 4.20 Torsion ofthin walled sections

and the other to warping torsion from the effects of the restraints. The proportions ofthe two components depend on the type of loading and the distance from a restraint.

Both components of torsion produce shear stresses parallel to the walls of the section,and their combined effects can locally be greater than the effect of St. Venant alone.However in members other than channels with very thin walls the increase in shearstresses can usually be ignored in design. In beams the maximum shear stress can beobtained approximately by combining the effects of transverse shears, using Eq. (4.15),with the shear stresses from torsion, assuming that the whole of the applied torqueresults in St. Venant torsion.

A more significant effect of warping restraint in the design of beams is the introductionof longitudinal stresses. The effect is illustrated for an ‘I’ beam in Fig. 4.20(b). Inthis case the warping displacements are confirmed to the flanges, whose positions, ifwarping were allowed to occur freely, are shown by dotted outlines. Bi-moments Mb

are induced in the planes of the flanges when warping is restrained, and these give riseto tensile and compressive stresses as shown.

A full treatment of the analysis of members subject to warping torsion is beyondthe scope of this book and the reader is referred to Zbirohowski-Koscia (1967). Theanalysis is tedious, but for ‘I’ beams a conservative estimate of the longitudinal stressesdue to warping torsion can be obtained by assuming that each flange acts independentlyand is bent in its own plane by an analogous system of lateral loads which replace theapplied torques, as in Fig. 4.20(c). The value of the equal and opposite lateral loadsH , analogous to the applied torque Mx is

H = Mx

h − tf(4.21)

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The ends of the flanges can be assumed to be either fixed or simply supported,depending on whether or not warping is restrained by the structural connections.The results obtained by this method are conservative because in reality the warp-ing stresses are produced only by the warping component of the applied torque,not the whole torque as assumed. However, Eq. (4.21) can be useful in preliminarydesigns where it is necessary to assess whether the effects of torsion are likely to besignificant.

4.4.3 Elastic Torsion of Circular Sections

The elastic (St. Venant) theory of torsion of prismatic members with solid or hollowsections can be expressed by the well established formula

TIt

= v

r= Gθ

L(4.22)

where

T is the torsional moment

It is the torsion constant which, for a circular section, is equal to the polar secondmoment of area

v is the shear stress at radius r

G is the shear modulus

θ is the angle of twist

L is the length of the member

The polar second moment of area for a solid section of radius R is

It = πR4

2(4.23)

For a thin walled tube of mean radius Rm and wall thickness t, an approximateformula is

It = 2πR3mt (4.24)

The error is below 3% for t/Rm ratios of 1/3 or less, and is on the safe side. The polarsecond moment of area (It) is twice the second moment of area (I) about a diameter,values for which are given in Section Tables.

The elastic distribution of shear stress along the radius of a solid circular section islinear, with zero at the centre and a maximum at the outside surface. For a thin walledtube, the stress varies linearly across the wall thickness and in the range of standardstructural tubes unsafe errors in the shear stress of up to about 18% are introducedby the use of Rm in Eq. (4.22) instead of the outside radius.

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4.4.4 Elastic Torsion of Thin Walled Open Sections

The torsional constant It for a thin rectangle of width b and thickness t is

It = bt3

3(4.25)

This formula is accurate when b/t is infinite and gives unsafe errors of 6% when b/t = 10,and 10% when b/t = 6. The b/t ratios are typical of the flanges of Universal Columnand Beam sections.

Most sections in steelwork design are composed of thin rectangles, and for a completesection the torsional constant can be obtained by summing the torsional constants foreach rectangular element, that is

It =∑

bt3

3(4.26)

Values of It are given in Section Tables. In standard rolled sections the root fillets atthe junctions of the web and the flanges give additional torsional stiffness.

The shear stresses in an open section under St. Venant torsion vary from zero on thecentre line of the wall to a maximum on the outside surface (Fig. 4.20(d)), and theirdirection is reversed on each side of the centre line. The shear flow constitutes a closedloop. The maximum stress in any element of thickness t is

v = TtIt

(4.27)

The maximum shear stress in a section therefore occurs in the thickest element. Atre-entrant corners the flow lines are crowded together, giving rise to very high stressconcentrations. The effect is reduced by fillet radii. The shear stress is zero at theoutside corners.

The angle of twist

θ = TLGIt

(4.28)

where It is calculated from Eq. (4.26).

4.4.5 Elastic Torsion of Thin Walled Closed Sections

The shear stress distribution for closed sections is shown in Fig. 4.20(e). The flow isunidirectional with respect to the profile, contrasting with open sections. Variationsin stress across the thickness of the section are ignored. The shear flow is constant atall points on the profile and is given by

v = T2A

(4.29)

where A is the cross-sectional area of the profile.

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The shear stress is a maximum in the thinnest part of the section and is obtained bydividing the shear flow by the thickness. This is in direct contrast to the open sectionwhere maximum shear stress occurs in the thickest part.

Angle of twist

θ = TL4A2G

∫dst

(4.30)

As in open sections additional stresses are introduced when warping is restrained.Equations (4.29) and (4.30) are derived from the Bredt–Batho hypothesis in which it isassumed that the shape of the section remains unchanged. To ensure that this assump-tion remains valid it may be necessary to stiffen the section with internal diaphragmsat intervals along its length, and at points where concentrated loads are applied.

EXAMPLE 4.7 Torsion and transverse shear in a box section. Find the maxi-mum shear stress in the box section shown in Fig. 4.21 which is subject to a torqueT = 200 kNm and a shear force V = 500 kN. Assume that the section is adequately stiff-ened to prevent distortion of the profile and ignore the effects of warping restraints.Calculate the angle of twist per metre length.

The torsion and shear may be considered separately and the resulting shear flows areshown in Figs 4.21(a) and (b). For both calculations the profile dimensions shown inFig. 4.21(c) can be used. The shear modulus is 80 GPa.

First consider torsion. The area enclosed by the profile

A = 790 × 380 = 0,3002E6 mm2.

From Eq. (4.29) the shear flow

v = T2A

= 200E62 × 0,3002E6

= 333 N/mm

The shear stress is maximum in the web where the section is thinnest

vt = v

t= 333

10= 33,3 MPa

380

790500 kN

T 400

800

20

10

T � 200 kNm

(a) Torsion (b) Shear (c) Profile

FIGURE 4.21Example: torsionand transverseshear in a boxsection

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From Eq. (4.30) the angle of twist

θ = TL4A2G

∫dst

= 200E6 × 1E3(4 × 0,30022 × 1E12 × 80 × 1E3)

× 2(

79020

+ 38010

)= 0,00107 rad/m

This is small which shows that the box is very stiff torsionally.

Now consider the direct shear force. The maximum shear stress is in the webs at theneutral axis, and the first moment of areas

Azc(flange) = 20 × 790 × 3802

= 3,002E6 mm3

Azc(webs) = 2 × 10 × 3802

× 3804

= 0,361E6 mm3

Total = 3,363E6 mm3

Second moment of area

I = 2

(10 × 3803

12+ 20 × 790 × 1902

)= 1232E6 mm4

Direct shear stress from Eq. (4.15)

vs = VAzc

It= 500E3 × 3,363E6

1232E6 × 2 × 10= 68,2 MPa

Combing torsional and direct shear stresses

v(combined) = vt + vs = 33,3 + 68,2 = 101,5 MPa

Information on combining shear and torsion in design calculations is given in cl 6.2.7,EN 1993-1-1 (2005).

4.5 PLASTIC SECTION PROPERTIES AND ANALYSIS

4.5.1 Plastic Section Modulus

Plastic global analysis may be used in the design of steel structures for class 1 cross-sections (cl 5.6, EN 1993-1-1 (2005)). For the general case of a steel section symmetricalabout the plane of bending, the stress distributions in the elastic and fully plastic stateare shown in Fig. 4.3(c). For equilibrium of normal forces, the tensile and compressiveforces must be equal. In the elastic state, when the bending stress varies from zero atthe neutral axis to a maximum at the extreme fibres, this condition is achieved whenthe neutral axis passes through the centroid of the section. In the fully plastic state,

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A4/2

A4/2

y4

A3

y3 � tw/4A1

A2

z 2

z 1

z

z

b f � 420,5

t f �

36,

6

tw � 21,5

h �

920

,5

yy

(a) (b) (c)

FIGURE 4.22 Example: plastic section modulus for an ‘I’ section

because the stress is uniformly equal to the yield stress, equilibrium is obtained whenthe neutral axis divides the section into two equal areas.

Mpl = (first moment of area about the plastic NA)fy (4.31)

EXAMPLE 4.8 Plastic section moduli for an ‘I’ section. Determine the plastic sectionmoduli about the y–y and z–z axes for the ‘I’ section shown in Fig. 4.22(a). The sectionis for a 914 × 419 × 388 kg Universal Beam with the root radius omitted.

To determine the plastic section modulus about the y–y axis divide the section into A1

and A2 as shown in Fig. 4.22(b) where

A1 =(

h2

)tw =

(920,5

2

)21,5 = 9895,375 mm2

A2 = (bf − tw)tf = (420,5 − 21,5)36,6 = 14603,5 mm2

and

z1 = h4

= 920,54

= 230,125 mm

z2 = h2

− tf2

= 920,5 − 36,62

= 441,95 mm

Plastic section modulus

Wply = 2(A1z1 + A2z2) = 2(9895,375 × 230,125 + 14603,4 × 441,95)

= 17,462E6 mm3

The value obtained from Section Tables is 17,657E6 mm3 which is slightly greaterbecause of the additional material at the root radius.

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Similarly for the plastic section modulus about the z–z axis divide the section into areasA3 and A4 as shown is Fig. 4.22(c) where

A3 = (h − 2tf )tw2

= (920,5 − 2 × 36,6)21,52

= 9108,475 mm2

A4 = 2(

bf

2

)tf = 2

(420,5

2

)36,6 = 15390,3 mm2

and

y3 = tw4

= 21,54

= 5,375 mm

y4 = bf

4= 420,5

4= 105,125 mm


Wplz = 2(A3y3 + A4y4) = 2(9108,475 × 5,375 + 15390,3 × 105,125)

= 3,334E6 mm3.

The value obtained from Section Tables is 3,339E6 mm3 which is slightly greaterbecause of the additional material at the root radius.

From Section Tables the ratio of plastic/elastic section modulus (shape factor) for thissection about the x–x axis is

Wply/Wely = 17,657E6/15,616E6 = 1,1307

The value of the shape factor for bending about the y–y axis generally quoted for‘I’ sections in current use in design is 1,15.

The corresponding value of the shape factor for bending about the z–z axis isWplz/Welz = 3,339E6/2,160E6 = 1,55 which is typical for an ‘I’ section.

EXAMPLE 4.9 Plastic section moduli for a channel section. Determine the plasticsection moduli about the y–y and z–z axes for the channel section shown in Fig. 4.23(a).The section is for a 432 × 102 × 65,54 kg channel with the root radius omitted.

To determine the plastic section modulus about the y–y axis divide the section intoA1 and A2 and shown in Fig. 4.23(b) where

A1 =(

h2

)tw =

(431,8

2

)12,2 = 2633,98 mm2

A2 = (bf − tw)tf = (101,6 − 12,2)16,8 = 1501,92 mm2

and

z1 = h4

= 431,84

= 107,95 mm

z2 = h2

− tf2

= 431,8 − 16,82

= 207,5 mm

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A1

A2

z 2

z 1

z 5

z 4

A5

A3

A4

yn

bf � 101,6

t f �

16,

8

tw � 12,2

h �

431

,8y y

(a) (b) (c)

FIGURE 4.23 Example:plastic section modulusfor a channel section


Wply = 2(A1z1 + A2z2) = 2(2633,98 × 107,95 + 1501,92 × 207,5)

= 1,192E6 mm3.

The value obtained from Section Tables is 1,207E6 mm3 which is slightly differentbecause of the additional material at the root radius and the fact that the flangestaper.

Similarly for the plastic section modulus about the z–z axis divide the section into areasA3, A4 and A5 and shown in Fig. 4.23(c). It is first necessary to determine the positionof the neutral axis z–z.

Since the axis z–z divides the total area into two equal parts

yn = A1 + A2

h= 2633,98 + 1501,92

431,8= 9,578 mm

Areas

A3 =(

h2

)yn =

(431,8

2

)9,578 = 2067,89 mm2

A4 =(

h2

− tf

)(tw − yn) =

(431,8

2− 16,8

)(12,2 − 9,578) = 522,04 mm2

A5 = (bf − yn)tf = (101,6 − 9,578)16,8 = 1545,97 mm2

and

y3 = yn

2= 9,578

2= 4,789 mm

y4 = tw − yn

2= 12,2 − 9,578

2= 1,311 mm

y5 = bf − yn

2= 101,6 − 9,578

2= 46,011 mm

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Wpz = 2(A3y3 + A4y4 + A5y5)

= 2(2067,89 × 4,789 + 522,04 × 1,311 + 1545,97 × 46,011)

= 0,1634E6 mm3

The value obtained from Section Tables is 0,1531E6 mm3 which is slightly differentbecause of the additional material at the root radius and the taper on the flanges.

4.5.2 Plastic Methods of Analysis

A plastic collapse mechanism depends on the formation of a plastic hinge(s) and thiswill now be considered in detail. The tensile stress–strain curve for mild steel is shownin Fig. 4.3(a). The curve is idealized into three stages namely elastic, plastic, andstrain hardening stages of deformation. The moment–curvature for a beam made ofthe same material is shown in Fig. 4.3(b) with the corresponding distribution of stressat various loading stages shown in Fig. 4.3(c). The spread of the plastic hinge alongthe length of the beam is shown in Fig. 4.3(d).

The amount of rotation that can take place at a plastic hinge is determined by the lengthof the yield plateau shown in Fig. 4.3(b). For mild steel the length is considerable andthe work hardening stage is ignored. For higher grade steels work hardening occursimmediately after yielding and there is no plateau. A plastic hinge does form but withan increasing moment of resistance. In design this increase in resistance due to workhardening is ignored which errs on the side of safety.

Plastic methods of analysis and design consider a structure at collapse when sufficientplastic hinges have formed to produce a mechanism. Examples of collapse mechanismsare shown in Fig. 4.24.

In simple situations, are shown in Fig. 4.24(a), the position of the plastic hinge isobvious and it is simple to calculate the collapse load using the method of virtual work.

L Pin

PinPinPlastic hingeθ θ1 θ2

Plastichinge

x

q /unit length q/unit length

(a) (b) (c)

L

FIGURE 4.24 Examples of plastic collapse mechanisms

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The method must only be applied to structures where the material becomes plasticat the yield stress and is capable of accommodating large plastic deformations. It musttherefore not be applied to brittle materials such as cast iron. However it can be appliedto reinforced concrete because the steel reinforcement behaves plastically at collapsebut care must taken to check the rotation at the hinge for heavily reinforced sections.

The plastic method can be seen as a more rational method for design because all partsof the structure can be given the same safety factor against collapse. In contrast forelastic methods the safety factor varies. Intrinsically the plastic method of analysisis simpler than the elastic method because there is no need to satisfy elastic straincompatibilty conditions. However calculations for instability and elastic deflectionsrequire careful consideration when using the plastic method, but nevertheless it isvery popular for the design of some structures (e.g. beams and portal frames).

The method of analysis demonstrated in this chapter is based on the principle ofvirtual work. This states that if a structure, which is equilibrium, is given a set of smalldisplacements then the work done by the external loads on the external displacementsis equal to the work done by the internal forces on the internal displacements. Moreconcisely, external work equals internal work. The displacements need not be real, theycan be arbitrary, which explains the use of the word ‘virtual’. However the externaland internal geometry must be compatible.

It is tacitly assumed that collapse is due to the formation of plastic hinges at cer-tain locations and that other possible causes of failure, for example, local or generalinstability, axial or shear forces, are prevented from occurring. It is also important tounderstand that at collapse:

(a) the structure is in equilibrium, that is, the forces and moments, externally andinternally, balance,

(b) no bending moment exceeds the plastic moment of resistance of a member,(c) there are sufficient hinges to form a collapse mechanism.

These three conditions lead to three theorems for plastic analysis.

(1) Lower bound theorem: if only conditions (a) and (b) are satisfied then the solutionis less than or equal to the collapse load.

(2) Upper bound theorem: if only conditions (a) and (c) are satisfied then solution isgreater than or equal to the collapse load.

(3) Uniqueness theorem: if conditions (a), (b) and (c) are satisfied then the solution isequal to the collapse load.

Settlement of the supports has no effect on the solution at collapse because the onlyeffect is to change the amount of rotation required. This is in contrast to elastic methodsof analysis where settlement calculations must be included.

Plastic hinges form in a member at the maximum bending moment. However at theintersection of two members, where the bending moment is the same, the hinge forms

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in the weaker member. Generally the locations of hinges are at restrained ends, inter-section of members and at point loads. The hinges may not form simultaneouslyas loading increases but this is not important for calculating the final collapse load.Generally the number of plastic hinges

n = r + 1

where

r is the number of redundancies.

However there are exceptions, for example, partial collapse of a beam in a structure.

Equating external to internal work for the member shown in Fig. 4.24(a)

wL(

L4

)θ = Mpl(2θ) hence Mpl = wL2/8

In other conditions, as shown in Fig. 4.24(c), the position of the plastic hinge forminimum collapse load is not obvious and the calculations are more complicated.Equating external to internal work for the member shown in Fig. 4.24(c).

qx(

xθ1

2

)+ q(L − x)2 θ2

2= Mpl(2θ1 + θ2) (i)

From geometry

xθ1 = (L − x)θ2 (ii)

Combining (i) and (ii)

qL = 2Mpl

(2x

− 1L − x

)(iii)

Differentiating (iii) to determine the value of x for which q is a minimum

x2 − 4Lx + 2L2 = 0 hence x = L(2 − √2) (iv)

Combining (iii) and (iv)

Mpl = (1,5 − √2)qL2 (v)

The method can be applied to a variety of structures. Further explanation and examplesare given in Moy (1981), and Croxton and Martin Vol. 2 (1987 and 1989).

4.6 EFFECT OF A SHEAR FORCE ON THE PLASTIC MOMENT OFRESISTANCE (CLS 5.6 AND 6.2.6, EN 1993-1-1 (2005))

In general the effect of a transverse shear force is to reduce the plastic moment ofresistance but the reduction for an ‘I’ section is small and may be ignored (cl 6.2.8, EN1993-1-1 (2005)) if

VEd

Vpl,Rd≤ 0,5 (4.32)

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where VEd is the design shear force, and the plastic shear resistance and (Eq. (6.18),EN 1993-1-1 (2005))

Vpl,Rd = Av

(fy/31/2)γM0

(4.33)

The areas resisting shear (Av) for various sections are given in cl 6.2.6(3), EN 1993-1-1(2005) and in Section Tables.

The area resisting shear for an ‘I’ section is

Av = A − 2btf + (tw + 2r)tf (4.34)

This is a slight increase on the web area (htw) which has been used in the past.

The maximum shear stress of fy/31/2 is based on the failure criterion expressed inEq. (2.5).

There is no reduction in the plastic moment of resistance for plastic or compact sec-tions provided that the design shear force does not exceed 50% of the plastic shearresistance. This recommendation is related to the work of Morris and Randall (1979)who stated that shear can be ignored unless the average shear stress in the web exceedsfy/3, or fy/4 when the ratio of the overall depth to flange width (h/bf ratio) exceeds 2.5.

Where the design shear force exceeds 50% of the plastic shear resistance the EuropeanCode recommends that the yield strength is reduced to (1 − ρ)fy where for shear(Eq. (6.29), EN 1993-1-1 (2005))

ρ =(

2VEd

Vpl,Rd− 1)2

(4.35)

and for torsion

ρ =(

2VEd

Vpl,T,Rd− 1)2

(4.35a)

These recommendations may be compared with theory by Horne (1971). If a shearforce is applied to an ‘I’ section most of the shear force is resisted by the web. If itis assumed that all of the shear force is resisted by the web the effect on the plasticmoment of resistance is shown in Fig. 4.25(a). The outer fibres in bending are at yieldwhile the inner fibres are linear elastic. It is not possible to resist shear forces on theouter fibres and thus the inner fibres resist all of the shear force.

Plastic moment of resistance of the web

Mplw =[

twd2

4− tw

d2p

4+ tw

d2p

6

]fy =

[tw

d2

4− tw

d2p

12

]fy (i)

From the parabolic distribution of shear stress

V = 23

twdpτy (ii)

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Parabolicdistributionof shear stress

h d tw dp

fy

τy

(a) 0 � V/Vmax � 2/3

(b) 2/3 � V /Vmax � 1

τy

fy

τ

σ

(c)

Eq. (4.37)

Eq. (4.36)

V /Vmax

1,00,80,60,40,20

1,0

0,8

0,6

0,4

0,2

Fact

or

FIGURE 4.25 Effect of a shear force on the plastic section modulus of the web of an‘I’ section

Defining

Vmax = twhτy (iii)

Combining (i) to (iii)

Mplw =(

twd2fy4

)[1 −

(34

)(V

Vmax

)2]

(4.36)

This expression is valid for 0 ≤ V/Vmax ≤ 2/3.

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When 2/3 ≤ V/Vmax ≤ 1 then assuming the stress distributions shown in Fig. 4.25(b)the plastic moment of resistance of the web

Mplw = twd2σ

6(i)

Applied shear force

V = dtwτ + 23

dtw(τy − τ) (ii)

If Vmax = dtwτy, ratio

V/Vmax = 13

(2 + τ

τ y

)(iii)

Adopting a failure criterion of the form shown Eq. (2.5)(σ

fy

)2

+(

τ

τy

)2

= 1 (iv)

Combining (i) to (iv)

Mplw = twd2fy4

×⎡⎣2

3

√1 −

(3V

Vmax− 2)2⎤⎦ (4.37)

The expressions in the square brackets in Eqs (4.36) and (4.37) are plotted inFig. 4.25(c) expressed as Mplw = factor(twd2fy/4). This theory is approximate and con-servative but it does give a general appreciation of the effect of a shear force on theplastic moment of resistance. A fuller description and less conservative theories aregiven in Horne (1958).

In addition when the ratio hw/tw > 72ε/η (cl 6.2.6, EN 1993-1-1 (2005)) the web shouldbe checked for shear buckling. The limiting value of hw/tw is related to experimentalwork by Longbottom and Heyman (1956), and later work by Horne (1958).

EXAMPLE 4.10 Plastic section modulus reduced by shear. Determine the plas-tic modulus for a 762 × 267 × 197 kg UB grade S275, using (a) Horne’s method, (b)European Code method, when subject to a design value of shear force VEd = 1145 kN.Ignore the material factor for this example.

(a) Horne’s method:

VEd

Vmax= VEd

htwfy/31/2 = 1145E3769,6 × 15,6 × 275/31/2 = 0,6

According to Horne, dividing Eq. (4.36) by fy

Wpl(web) = td2

4

[1 − 3

4

(VEd

Vmax

)2]

= 0,73(td2/4), that is 27% loss in the web.

Ch04-H5060.tex 11/7/2007 17: 5 page 77


Reduced plastic section modulus of the complete section

Wpl(reduced) = Wpl(whole) − Wpl(web)(1 − factor)

= 7,164E6 −(

15,6 × 685,82

4

)(1 − 0,73) = 6,672E6 mm3

Percentage reduction in plastic modulus for the whole section

= 100(7,167E6 − 6,672E6)7,167E6

= 6,91%

(b) EN 1993-1-1 (2005) method:

VEd

Vpl,Rd= VEd

(Avfy/31/2)

= 1145E3(12,7E3 × 275/31/2)

= 0,568

where Av = 12,7E3 is obtained from Section Tables.

According to cl 6.29, EN 1993-1-1 (2005) there is a reduction in the plasticmodulus if

VEd

Vpl,Rd> 0,5

From Eq. (4.35a)

ρ =(

2VEd

Vpl,Rd− 1)2

= (2 × 0,568 − 1)2 = 0,0185

and

Wpl(reduced) = Mv,Rd

fy= Wpl − ρA2

v

4tw

= 7,164E6 − 0,0185 × 12,7E32

4 × 15,6

= 7,116E6 mm3

Percentage reduction = (7,164E6 − 7,116E6) × 1007,164E6

= 0,67%.

4.7 LATERAL RESTRAINT (CL 6.3.5, EN 1993-1-1 (2005))The full rotation required at a plastic hinge in a beam may not be realized unless lateralsupport is provided at the hinge position. It may also be necessary to provide lateralsupport at other points along the span to ensure that lateral torsional buckling doesnot occur. Lateral torsional buckling is considered to be prevented if the compressionflange is prevented from moving laterally, either by an intersecting member, or byfrictional restraint from intersecting floor units.

Ch04-H5060.tex 11/7/2007 17: 5 page 78


4.8 RESISTANCE OF BEAMS TO TRANSVERSE FORCES

The first check for transverse forces is the shear stress in the web at the neutral axisin the elastic stage of behaviour (cl 6.2.6(4), EN 1993-1-1 (2005)). In the plastic stageof behaviour the strength of the web is determined using values given in cl 6.2.6(2),EN 1993-1-1 (2005). Also it is necessary to consider web plate shear buckling at theultimate limit state by checking if 72ε/η > hw/tw (cl 5.1(2), EN 1993-1-5(2003)).

Design calculations are also required for concentrated transverse forces applied togirders from supports, cross beams, columns, etc. (Fig. 4.26). The concentrated loadsare dispersed through plates, angles and flanges to the web of the supporting girder.The deformations that occur to the supporting beam are shown in Fig. 4.27 and includeyielding of the flange and local buckling of the web as shown in experiments (Astillet al. (1980)).

The design resistance is expressed simply as (cl 6.2, Eq. (6.1), EN 1993-1-5(2003))

FRd = fywLeff twγM1

The effective bearing length (Leff ) is an extension of the stiff bearing length (ss) whichassumes a 45◦ dispersion through plates, flanges and angles as shown in Fig. 4.26.The root radius of a section increases the length of the stiff bearing by (2 − 21/2)r.

‘I’ beam

45°

t p

hss = h + 2tp

ss

(a) (b)

(c)

‘I’ beam

rb

t f

‘I’ beam

45°ss = 2t f + tw + 2(2 − √2 )rb

ss

tw

Clearance

Angle support

ta

ra

‘I’ beam

45°

Ss

ss = 2ta + (2 − √2 )ra − Clearance

FIGURE 4.26 Stiff bearing lengths

Ch04-H5060.tex 11/7/2007 17: 5 page 79


Buckling of web

tw

Yielding of flange

t fLeff

ss

FIGURE 4.27 Transverse concentrated load

The extension of the stiff bearing length is based on theory of flange yielding andbuckling of the web as shown in Examples 4.11 and 4.12.

If the transverse resistance of an unstiffened web is insufficient stiffeners are required(cl 9.4(1), EN 1993-1-5 (2003)) designed according to cl 6.3.3 or 6.3.4, EN 1993-1-1(2005) with a buckling length of not less than 0,75hw and using buckling curve c (Fig. 6.4,EN 1993-1-1 (2005)).

EXAMPLE 4.11 Simply supported beam carrying a uniformly distributed load andlaterally restrained. The floor of an office building consists of 125 mm precast concreteunits, with a mass of 205 kg/m2, topped with a 40 mm concrete screed and 20 mm woodblocks. Lightweight partitions supported by the floor are equivalent to a superficialload of 1,0 kN/m2 and the suspended ceiling has a mass of 40 kg/m2. The floor rests onthe top flanges of simply supported steel beams spanning 8 m and at a pitch of 3,75 m.

Characteristic loads kg/m2 kN/m2

Dead loadSelf-weight of steel beam (assumed) 20125 mm precast units 20540 mm concrete screed 2400 × 0,04 9620 mm wood blocks 900 × 0,02 18Suspended ceiling 40

379 × 9,81/1E3 = 3,72Lightweight partitions 1,00

Total dead load 4,72Imposed load for an office (EN 1991-1-1 (2002)) 2,50

Maximum design bending moment at mid-span kNm

Permanent load BM = γGQL/8 = 1,35 × 4,72 × 82 × 3,75/8 = 191,2Variable load BM = γQQL/8 = 1,5 × 2,5 × 82 × 3,75/8 = 112,5

Total BM 303,7

Ch04-H5060.tex 11/7/2007 17: 5 page 80


Using grade S275 steel, fy = 275 MPa, the plastic section modulus required

Wply = Mfy/γM

= 303,7E6275/1,1

= 1,215E6 mm3.

From Table 4.1 for L/250, L/h = 22, hence h = 8E3/22 = 364 mm.

From Section Tables, try 457 × 152 × 60 kg UB, Wply = 1,284E6 mm3,

tf = 13,3 mm, tw = 8 mm, d = 407,7 mm, Av = 3890 mm2, bf = 152,9 mm,

h = 454,7 mm, hw = (h − 2tf ) = (454,7 − 2 × 13,3) = 428,1 mm,

r = 10,2 mm.

Check for buckling of web (Table 5.2 (sheet 1), EN 1993-1-1 (2005))

dtw

= 407,88

= 50,96

72ε = 72(

235fy

)1/2

= 72(

235275

)1/2

= 66,6 > 50,96 satisfactory.

Check flange buckling (Table 5.2 (sheet 2), EN 1993-1-1 (2005))

ctf

= [(B − tw)/2 − r]tf

= [(152,9 − 8)/2 − 10,2]13,3

= 4,67

9ε = 9 ×(

235fy

)1/2

= 9(

235275

)1/2


This is a Class 1 section and calculations can be reduced by using Section Tables.

Check deflection at the service limit state

Permanent and variable load on span = 2,5 × 8 × 3,75 = 75 kN

Maximum deflection = δ1 + δ2 = 5QL3

384EI

= 5 × 75 × (8E3)3

384 × 210 × 254,64E6= 9,35 mm

Deflection limit from Table 4.1

δmax = L250

= 8E3250

= 32 > 9,35 mm satisfactory.

Check shear resistance of web at the ultimate limit state.

Design shear force at the support

VEd = QL2

= (1,35 × 4,72 + 1,5 × 2,5)8 × 3,752

= 151,8 kN.

Area of web (cl 6.2.6(3), EN 1993-1-1 (2005))

Av = A − 2btf + (tw + 2r)tf = 7580 − 2 × 152,9 × 13,3 + (8 + 2 × 10,2)13,3

= 3890 mm2, or obtain value from Section Tables.

Ch04-H5060.tex 11/7/2007 17: 5 page 81


Design plastic shear resistance (cl 6.2.6(1), EN 1993-1-1 (2005))

Vpl,Rd = Avτy

γM0= Av(fy/31/2)

γM0

= 3890 × (275/31/2)1,1 × 1E3

= 561,5 kN

VEd

Vpl,Rd= 151,8

561,5= 0,27 < 1,0 satisfactory.

Check web plate buckling from shear at the ultimate limit state (cl 5.1 (2), EN 1993-1-5(2003))

72ε

η= 72(235/275)1/2

1, 2= 55,5

hw

tw= 428,1

8= 53,5 < 55,5 satisfactory.

Assuming 150 × 75 × 10 mm angle supports at the ends of the beam (Fig. 4.26) thetransverse shear buckling load (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003))


= 275 × 77,53 × 81,0 × 1E3

= 170,6 > VEd = 151,8 kN satisfactory no stiffener required.

The effective length (Leff ) in the previous equation is obtained as follows:

The stiff bearing length (cl 6.3, Fig. 6.2, EN 1993-1-5 (2003)), for a 150 × 75 × 10 mmangle support (Fig. 4.26)

ss = 2ta + (2 − 21/2)ra − clearance

= 2 × 10 + (2 − 21/2) × 11 − 3 = 23,44 < hw = 428,1 mm

Buckling co-efficient for the load application (cl 6.1(4), Type(c), EN 1993-1-5 (2003))

kF = 2 + 6(ss + c)hw

= 2 + 6(23,44 + 0)428,1

= 2,33 < 6 satisfactory.

Effective load length (cl 6.5, Eq. (6.13), EN 1993-1-5 (2003))

le = kFEt2w

2fywhw

= 2,33 × 210E3 × 82

2 × 275 × 428,1= 133,0 > (ss + c = 23,44 + 0) use 23,44 mm.

Force (cl 6.4 Eq. (6.5) EN 1993-1-5 (2003))

Fcr = 0,9 kFEt3w

hw= 0,9 × 2,33 × 210E3 × 83

(428,1 × 1E3)= 526,7 kN.

Ch04-H5060.tex 11/7/2007 17: 5 page 82


The dimensionless parameters (cl 6.5, Eq. (6.8), EN 1993-1-5 (2003))

m1 = fyfbf

fywtw= 275 × 152,9

275 × 8= 19,11

and (cl 6.5, Eq. (6.9), EN 1993-1-5 (2003))

m2 = 0,02(

hw

tf

)2

= 0,02(

428,113,3

)2

= 20,72

Yield length (cl 6.5, Eq. (6.12), EN 1993-1-5 (2003))

ly = le + tf (m1 + m2)1/2 = 23,44 + 13,3(19,11 + 20,72)1/2 = 107,4 mm

An alternative yield length (cl 6.5, Eq. (6.11), EN 1993-1-5 (2003))

ly = le + tf

[m1

2+(

letf

)2

+ m2

]1/2

= 23,44 + 13,3

[19,11

2+(

23,4413,3

)2

+ 20,72

]1/2

= 100,3 mm use minimum.

An alternative yield length (cl 6.5, Eq. (6.10), EN 1993-1-5 (2003))

ly = ss + 2tf [1 + (m1 + m2)1/2]

= 23,44 + 2 × 13,3[1 + (19,11 + 20,72)1/2] = 217,9 mm.

Factor (cl 6.4, Eq. (6.4), EN 1993-1-5 (2003))

λF =(

lytwfyw

Fcr

)1/2

=(

100,3 × 8 × 275526,7E3

)1/2

= 0,647

Reduction factor (cl 6.4, Eq. (6.3), EN 1993-1-5 (2003))

χF = 0,5λF

= 0,50,647

= 0,773

The effective length (cl 6.2, Eq. (6.2), EN 1993-1-5 (2003))

Leff = χFly = 0,773 × 100,3 = 77,53 < hw = 428,1 mm.

Check self-weight of steel beam = 60/3,75 = 16 < 20 kg/m2 assumed in loading calcu-lations, acceptable.

EXAMPLE 4.12 Support for a conveyor. Part of the support for a conveyor consistsof a pair of identical beams as shown in Fig. 4.28. Each beam is connected to a stanchionat end A by a cleat and is supported on a cross beam at D by bolting through theconnecting flanges. Lateral restraint is provided by transverse beams at A, B and Econnected to rigid supports. The loads shown are at the ultimate limit state.

Ch04-H5060.tex 11/7/2007 17: 5 page 83


600

300

375

75 150

825

300

Shear force (kN)

2 m3 m3 m2 m

533 × 210 × 101 kg UB

203 × 203 × 46 kg UC

A B C D E

225 kN150 kN

450 kN

15

150 × 75 × 10 mm angle610 × 229 × 140 kg UB


FIGURE 4.28 Example: support for a conveyor

Assuming pin joints at A and D.

Reactions are determined by taking moments about A, then about D.

RD = 2 × 225 + 5 × 450 + 10 × 1508

= 525 kN

RA = 6 × 225 + 3 × 450 − 2 × 1508

= 300 kN

Important bending moments are

MB = RA × 2 = 300 × 2 = 600 kNm

MC = RA × 5 − 225 × 3 = 300 × 5 − 225 × 3 = 825 kNm

MD = −150 × 2 = −300 kNm

The shear force and bending moment diagrams at the ultimate limit state are shownin Fig. 4.28.

Ch04-H5060.tex 11/7/2007 17: 5 page 84


Using Grade S355 steel with a characteristic strength fy = 355 MPa. The plastic sectionmodulus required

Wply = Mmax

fy/γM0= 825E6

355/1,1= 2,56E6 mm3.

Deflection limits (Table 4.1), L/250, L/h = 17, hence h = 8E3/17 = 470 mm.

From Section Tables try 533 × 210 × 101 kg UB, a Class 1 section,

Wply = 2,619E6 mm3, tf = 17,4 mm, tw = 10,9 mm, Av = 6250 mm2, d = 476,5 mm.

Check deflections at service load

Using area and area moment methods the deflections due to the imposed loads atservice loads are:

mid-span in ABCD 11,3 mm (downwards)end of cantilever DE −6,5 mm (upwards)

Deflection limits (Table 4.1)

span ABCD = L250

= 8E3250


span DE = 2L250

= 2 × 2E3250


Check web shear resistance between C and D at the ultimate limit state.

Design shear force VEd = 375 kN.

Plastic shear resistance Vpl,Rd = Av( fy/31/2)γM

= 6240 × (355/31/2)1,1 × 1E3

= 1163 > 375 kN satisfactory.

Shear area (cl 6.2.6, EN 1993-1-1 (2005))

Av = A − 2btf + (tw + 2r)tf

= 129,3 × 100 − 2 × 210,1 × 17,4 + (10,9 + 2 × 12,7) × 17,4

= 6250 mm2 or obtain a value from Section Tables.

VEd/Vpl,Rd = 375/1163 = 0,322 < 0,5 therefore no reduction in the plastic sectionmodulus (cl 6.2.8, EN 1993-1-1 (2005)).

Check for buckling of web (Table 5.2 (sheet 1), EN 1993-1-1 (2005))

ctw

= 476,510,9

= 43,7

72ε = 72(

235fy

)1/2

= 72(

235355

)1/2


Ch04-H5060.tex 11/7/2007 17: 5 page 85



ctf

= [(B − tw)/2 − r]tf

= [(210,1 − 10,9)/2 − 12,7]17,4

= 4,99

9ε = 9 ×(

235fy

)1/2

= 9 ×(

235355

)1/2



Assuming a 150 × 75 × 10 mm angle support at A at the end of the beam (Fig. 4.26) thetransverse shear buckling strength (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003)) is determinedas shown previously in Example 4.11.

At D (Fig. 4.28) two UBs intersect which have the following dimensions.

From Section Tables the 533 × 210 × 101 kg upper load carrying beam ABCDE.h = 536,7 mm, bf = 210,1 mm, hw = (h − 2tf ) = (536,7 − 2 × 17,4) = 501,9 mm,tf =17,4 mm, tw = 10,9 mm, r = 12,7 mm.

From Section Tables the 610 × 229 × 140 kg lower support beam at D.h = 617 mm, bf = 230,1 mm, hw = (h − 2tf ) = (617 − 2 × 22,1) = 572,8 mm, tf =22,1 mm, tw =13,1 mm, r = 12,7 mm.

The transverse shear buckling strength of the upper 533 × 210 × 101 kg load carryingbeam (cl 6.2, Eq. (6.1), EN 1993-1-5 (2003)).


= 355 × 501,9 × 10,91,0 × 1E3

= 1942 > RD = 525 kN satisfactory no stiffener required.

The previous calculation for FRd includes the effective bearing length (Leff ) which iscalculated as follows:

The stiff bearing length (cl 6.3, Fig. 6.2, EN 1993-1-5 (2003)), provided by the flangeand web of the lower 610 × 229 × 140 UB

ss = 2tf + tw + (2 − 21/2)rb

= 2 × 22,1 + 13,1 + (2 − 21/2) × 12,7 = 64,74 < hw = 501,9 mm

Buckling co-efficient for the load application (cl 6.1(4), Type (a), EN 1993-1-5 (2003))assuming a is large

kF = 6 + 2(

hw

a

)2

= 6

Force (cl 6.4, Eq. (6.5), EN 1993-1-5 (2003))

Fcr = 0,9 kFEt3w

hw= 0,9 × 6 × 210E3 × 10,93

501,9 × 1E3= 2926 kN

Ch04-H5060.tex 11/7/2007 17: 5 page 86


The dimensionless parameters (cl 6.5, Eq. (6.8), EN 1993-1-5 (2003))

m1 = fyfbf

fywtw= 355 × 210,1

355 × 10,9= 19,28

and (cl 6.5, Eq. (6.9), EN 1993-1-5 (2003))

m2 = 0,02(

hw

tf

)2

= 0,02(

501,917,4

)2

= 16,64

Yield length (cl 6.5, Eq. (6.10), EN 1993-1-5 (2003))

ly = ss + 2tf [1 + (m1 + m2)1/2]

= 64,74 + 2 × 17,4[1 + (19,28 + 16,64)1/2] = 308,1 mm

Factor (cl 6.4, Eq. (6.4), EN 1993-1-5 (2003))

λF =(

lytwfyw

Fcr

)1/2

=(

308,1 × 10,9 × 3552926E3

)1/2

= 0,638

Reduction factor (cl 6.4, Eq. (6.3), EN 1993-1-5 (2003))

χF = 0,5λF

= 0,50,638

= 0,784 < 1

The effective length (cl 6.2, Eq. (6.2), EN 1993-1-5 (2003))

Leff = χFly = 0,784 × 308,1 = 241,6 < hw = 501,9 mm

Similar calculations are required for the web buckling strength of the upper533 × 210 × 101 kg UB at B, C and E.

Stiffeners are not required for the 533 × 210 × 101 kg UB but in the past many design-ers have inserted them where large point loads are applied. Assuming that oneis required at D(RD = 525 kN) the following calculations are required. Assume asymmetrical stiffener of 10 mm thickness welded to the web and the flanges.

The non-dimensional slenderness ratio of the stiffener (Eq. (6.50), EN 1993-1-1(2005))

λz =(

AfyNcr

)1/2

= (Lcr/iy) × 1λ1

= (0,75 × 572,8/15E3) × 1[93,9 × (235/355)1/2]

= 0,375E-3

assuming a stiffener on both sides of the web, breadth of stiffener

bs = bf − tw − 2 (weld leg) = 210,1 − 10,9 − 2 × 6 = 187,2 (use 180) mm

iz = bst3s /12 = 180 × 103/12 = 15E3 mm4

Ch04-H5060.tex 11/7/2007 17: 5 page 87


q = 42,5 kN/m

Q85 kN

Q85 kN

CE

11 m 3 m 3 m 3 m

A B D


Collapse of span BCCollapse of span AB44

1

441

478,

1478,

1

190,3

329,4

201,9

223,1

95,610,6

277,2

116,9 Shear force (kN)

FIGURE 4.29 Example: two span beam

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve c) with λz = 0,375E-3 the reductionfactor χz = 1,0 and the design buckling resistance

Nb,Rd = χzAfyλM

= 1,0 × 180 × 10 × 3551,1 × 1E3

= 581 > RD = 525 kN satisfactory.

The design resistance of four vertical 6 mm fillet welds connecting the stiffener to bothsides of the web (cl 4.5.3.3(3), Eq. (4.4), EN 1993-1-8 (2005))

Fw,Rd = 4hw fu/(31/2βγM2) × 0,7a

= 4 × 572,8 × 510/(31/2 × 0,9 × 1,25) × 0,7 × 6/1E3

= 2519 > RD = 525 kN satisfactory.

EXAMPLE 4.13 Two span beam. Determine the size of Universal Beam requiredto support the design loads at the ultimate limit state as shown in Fig. 4.29. Assumethat the compression flange is fully restrained and that lateral torsional buckling doesnot occur.

Ch04-H5060.tex 11/7/2007 17: 5 page 88


Plastic analysis of the beam produces the following.

Collapse of span AB considered as a propped cantilever

Mpl =(

1,5 − √2)

qL2 =(

1,5 − √2)

42,5 × 112 = 441 kNm.

Collapse of span BC with plastic hinges assumed to be at E and B (Fig. 4.29)

External work = internal work

Q(

2L3

)θ1 + Q

(2L3

)θ1

2+ q

(2L3

)(L3

)θ1 + q

(L3

)(L6

)θ2

= Mpl(θ1 + θ1 + θ2) (i)

and from geometry(2L3

)θ1 =

(L3

)θ2 hence θ2 = 2θ1 (ii)

Combining eqs (i) and (ii) and rearranging

Mpl = QL4

+ qL2

12= 85 × 9

4+ 42,5 × 92

12= 478,1 kNm

This value of Mpl is the greater than the value for span AB and therefore is used todetermine the size of a section which is continuous for two spans. The assumptionthat a plastic hinge is at E is not correct, it actually occurs at 3,25 m from C andMpl = 479,1 kNm. However the error is small and is ignored.

Used Grade S355 steel and assuming a characteristic strength fy = 355 MPa the plasticsection modulus required

Wply = Mmax

fy/γM= 478,1E6

355/1,0= 1,347E6 mm3

From Section Tables try 457 × 191 × 67 kg UB which is a Class 1 section bendingabout the y–y axis, Wply = 1,471E6 mm3, tf = 12,7 mm, tw = 8,5 mm, Av = 4100 mm2,d = 407,8 mm, bf = 189,9 mm.

Check deflections at service load.

Using area and area moment methods the deflections due to the imposed load are20,7 mm for AB and 25 mm for BC.

Deflection limit (Table 4.1)

AB isL

250= 11E3

250= 44 > 20,7 mm, satisfactory

BC isL

250= 9E3

250= 36 > 25 mm, satisfactory.

Ch04-H5060.tex 11/7/2007 17: 5 page 89


Sketch the shear force diagram (Fig. 4.29) at the ultimate limit state for the collapseof span BC.

For EC

E

�

+ 478,1 + 42,5 × 3 × 1,5 − 3RC = 0 hence RC = 223,1 kN

For AB

B

�

+ 478,1 − 42,5 × 11 × 112

+ 11RA = 0 hence RA = 190,3 kN

For ABC

↑ + RA + RB + RC −∑

qL −∑

Q = 0

+190,3 + RB + 223,1 − 42,5 × 20 − 2 × 85 = 0 hence RB = 606,6 kN

Check web shear resistance at B at the ultimate limit state.

Design shear force

VEd = 329,4 kN (Fig. 4.29)

Design plastic shear resistance

Vpl,Rd = Avτy

γM= Av(fy/31/2)

γM= 4100 × (355/31/2)

1,0 × 1E3

= 840 > 329,4 kN satisfactory.

VEd/Vpl,Rd = 329,4/840 = 0,392 < 0,5 therefore no reduction in the plastic sectionmodulus (cl 6.2.8, EN 1993-1-1 (2005)).

Check for buckling of the web (Table 5.2 (sheet 1), EN 1993-1-1 (2005))

ctw

= 407,98,5

= 48,0

72ε = 72(

235fy

)1/2

= 72(

235355

)1/2



ctf

= [(B − tw)/2 − r]tf

= [(189,9 − 8,5)/2 − 10,2]12,7

= 6,33

9ε = 9 ×(

235fy

)1/2

= 9 ×(

235355

)1/2



Ch04-H5060.tex 11/7/2007 17: 5 page 90


A continuous 457 × 191 × 67 kg UB can be used for both spans. Alternatively if aminimum weight design is required then:

(a) A smaller section could be used for span AB but there would have to a splicewhich would increase fabrication costs.

(b) A smaller section could be used for span AB, with flange plates added to increasethe moment of resistance for span BC, but again this would increase fabricationcosts.

Calculations, similar to Example 4.12, are required for the web buckling strength ofthe 457 × 191 × 67 kg UB at A, B, and C.

REFERENCES

Astill, A.W., Holmes, M. and Martin, L.H. (1980). Web buckling of steel ‘I’ beams. CIRIA Tech.note 102.

Croxton, P.C.L. and Martin, L.H. (1987 and 1989). Solving problems in structures Vols. 1 and 2.Longman Scientific and Technical.

EN 1993-1-1 (2005). General rules and rules for buildings. BSI.EN 1993-1-5 (2003). Plated structural elements. BSI.EN 1993-1-8 (2005). Design of joints. BSI.Horne, M.R. (1971). Plastic theory of structures. Nelson.Horne, M.R. (1958). The full plastic moment of sections subjected to shear force and axial loads,

British Welding Journal, 5, 170.Longbottom, E. and Heyman, J. (1956). Experimental verification of the strength of plate girders designed

in accordance with the revised BS153: tests on full size and on model plate girders, Proceedings of theICE, 5(III), 462.

Megson, T.H.G. (1980). Strength of materials for civil engineers. Nelson.Morris L.J. and Randall A.L. (1979) Plastic Design, Constrado.Moy, S.S. (1981). Plastic methods for steel and concrete structures. Macmillan Press.Zbirohowski-Koscia, K. (1967). Thin walled beams. crosby Lockwood.

Ch05-H5060.tex 12/7/2007 15: 15 page 91

C h a p t e r 5 / Laterally UnrestrainedBeams

In the previous chapter it was assumed that the compression flange of a beam was fullyrestrained, that is, the flange is unable to move laterally under the effect of loads oractions. In practice, this is rarely the case and although compression flanges may berestrained at discrete points the flange is still capable of buckling between restraints.Such buckling reduces the moment capacity of the member.

5.1 LATERAL TORSIONAL BUCKLING OF ROLLED SECTIONS SYMMETRICABOUT BOTH AXES

5.1.1 Basic Theory

A beam under the action of flexure alone due to the application of point momentsproducing single curvature is considered as the basic case. The supports allow rotationabout a vertical axis but do not allow relative displacement of the top and bottomflanges, that is, twisting is not allowed (Fig. 5.1).

The governing equation for flexure about the minor axis is

EIzd2v

dx2 = −Mφ (5.1)

where EIz is the flexural rigidity about the minor axis.

The governing equation for torsion is given by

GItdφ

dx− EIw

d3φ

dx3 = Mdv

dx(5.2)

where GI t is the torsional rigidity, EIw the warping rigidity and the final term is thedisturbing torque.

Eliminate the term dv/dx between Eqs (5.1) and (5.2) to give

GItd2φ

dx2 − EIwd4φ

dx4 = −M2

EIzφ (5.3)

Ch05-H5060.tex 12/7/2007 15: 15 page 92

92 • Chapter 5 / Laterally Unrestrained Beams

Pins PinsM M

L

(a) Elevation of beam loaded in single curvature

Buckled position

Unbuckled position

(b) Support arrangement with f = d2f

dx2= 0

V

(c) Plan view of buckled beam

Original position

Buckled position

V

f

(d) Section

Beam afterbuckling

L

x

of unbuckledbeam

dvdx

cL

cL

FIGURE 5.1 Lateral torsional buckling of beams

Ch05-H5060.tex 12/7/2007 15: 15 page 93


The solution to Eq. (5.3) is given by the first term of a sine series

φ = φ0 sinπxL

(5.4)

where φ0 is the angle of twist at mid-span.

Substituting Eq. (5.4) into Eq. (5.3) with the boundary condition that φ = 0 at x = Lgives the elastic critical moment Mcr as

Mcr = π2EIz

L2

[Iw

Iz+ L2GIt

π2EIz

]1/2

(5.5)

or

Mcr = π√

EIzGIt

L

[1 + π2EIw

L2GIt

]1/2

(5.6)

The full derivation of the elastic critical moment is given in Kirby and Nethercot (1979)or Trahair and Bradford (1988).

In practice, real beams do not achieve full elastic buckling except at very high slen-derness ratios and are also subject to the limit imposed by the section plastic capacity.Reductions in elastic buckling capacity are caused by the existence of residual stressesand any lack of initial straightness. The residual stresses within the section arise inthe case of rolled sections due to differential rates of cooling in the web and flangesafter hot rolling, and of plate girders due to both the preparation of the web andflange plates and the welding procedure (Nethercot, 1974a). These residual stressesdo not affect the plastic capacity (Nethercot, 1974b; Kirby and Nethercot, 1979). Thevalue of the critical moment given by Eq. (5.5) or (5.6) takes no account of majoraxis flexure. Trahair and Bradford (1988) indicate the critical moment obtained fromEq. (5.5) or (5.6) should be divided by a factor K where K is given by

K =√(

1 − EIz

EIy

)(1 − GIt

EIy

[1 + π2EIw

GItL2

])(5.7)

In practice for I or H sections GIt/EIy[1 + π2EIw/GItL2] is negligible compared to

unity and thus K may be taken as

K =√

1 − EIz

EIy=√

1 − Iz

Iy(5.8)

For British Universal Beams the value of K is between 0,94 and 0,97 thus increasingMcr by around 3% to 6%. This enhancement may therefore be neglected for UniversalBeams. For Universal Columns, however, K is between 0,80 and 0,83, thus increasingMcr by around 17% to 20%, and should therefore possibly be taken into account.

Ch05-H5060.tex 12/7/2007 15: 15 page 94


0 0.4 0.8 1.2 1.6

Design curve

(Lower bound toexperimental results)

Experimentalresults

Elastic buckling curve (Eq.(5.5) orEq.(5.6))

Perfect beambehaviour

1.0

0.5Mb ,

Rd

MpI

Effective slenderness⎛⎜⎝

⎛⎜⎝

Mpl

Mcr

½

Full plastic action

FIGURE 5.2 Beam behaviour with lateral torsional buckling

5.1.2 Interaction between Plastic Moment Capacity, ElasticCritical Moment and Allowable Bending Moment

If the results for beam strength against effective slenderness are plotted in non-dimensional format the results appear as in Fig. 5.2. The performance of beams isreduced below the perfect conditions due to the residual stresses and initial imper-fections. A convenient lower bound to test data is given by a Perry–Robertson typeapproach to modelling the interaction. The imperfection co-efficient makes allowanceboth for geometric imperfections and residual stress levels. It should be noted thatunlike columns there is no theoretical justification for such an approach.

5.1.2.1 Design Interaction between Buckling and Plasticity

For any type of beam, the interaction equation is given by

(Mb,Rd − Mcr)(Mb,Rd − Mpl,Rd) = ηLTMcrMb,Rd (5.9)

where ηLT is the imperfection factor for lateral torsional buckling.

Equation (5.9) may be written in a normalized form to give(Mb,Rd

Mpl,Rd− Mcr

Mpl,Rd

)(Mb,Rd

Mpl,Rd− 1)

= ηLTMcr

Mpl,Rd

Mb,Rd

Mpl,Rd(5.10)

It is convenient to define two parameters χLT and λLT given as

χLT = Mb,Rd

Mpl,Rd(5.11)

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TABLE 5.1 Values of αLT for the generalized lateraltorsional buckling case.

Cross section Size limits (h/b) αLT

Rolled I sections ≤2 0,21>2 0,34

Welded I sections ≤2 0,49>2 0,76

Other cross-sections 0,76

λLT =[

Mpl,Rd

Mcr

]1/2

(5.12)

Equation (5.10) then becomes

(χLT − 1)(χLT − (λLT)2) = ηLTχLT

(λLT)2(5.13)

The solution to which is given by

χLT = 1�LT + [�2

LT − (λLT)2]1/2(5.14)

where �LT is given by

�LT = 0,5[1 + ηLT + (λLT)2] (5.15)

with ηLT given as

ηLT = αLT(λLT − 0,2) (5.16)

The values of αLT are dependant upon the type of section and whether rolled or welded.The appropriate values are given in Table 5.1 (cl 6.3. 2.1. EN 1993-1-1)

In design, full plastic moment capacity may be mobilized when λLT is less than λLT,0

which may be taken as 0,4 or when MEd/Mcr ≤ (λLT,0)2.

5.1.2.2 Rolled Sections and Equivalent Welded Sections (cl 6.3.2.3EN 1993-1-1)

For rolled sections and welded sections which are by implication symmetric, analternative formulation of the interaction formula may be used,

χLT = 1�LT + [�2

LT − β(λLT)2]1/2(5.17)

where �LT is given by

�LT = 0,5[1 + ηLT + β(λLT)2] (5.18)

with ηLT as

ηLT = αLT(λLT − λLT,0) (5.19)

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TABLE 5.2 Values of αLT for lateral torsional bucklingof rolled and welded sections.

Cross section Size limits (h/b) αLT

Rolled I sections ≤2 0,34>2 0,49

Welded I sections ≤2 0,49>2 0,76

The parameter χLT is subject to the limits χLT ≤ 1,0 and ≤ (λLT)2. The code recom-mends that the maximum value of λLT,0 should be 0,4 and the minimum value of β

as 0,75. The UK National Annex will adopt these values but will effectively specifythe approach should be limited to rolled sections, and that, therefore, the generalizedmethod must be used for welded sections.

Again the values of αLT are dependant upon the type of section and whether rolled orwelded. The appropriate values are given in Table 5.2.

Additionally, if this method is used, then the calculation of Mb,Rd needs modifying byusing a factor χLT,mod instead of χLT where

χLT,mod = χLT

f(5.20)

The factor f is given by

f = 1 − 0,5(1 − kc)[1 − 2,0(λLT − 0,8)2] ≤ 1,0 (5.21)

The factor kc is determined from the type of loading.

For loading only at points of restraint,

kc = 11,33 − 0,33ψ

(5.22)

where ψ is the ratio between the moments at restraint points subject to the conditionthat −1 ≤ ψ ≤ 1. For loading between restraints values of kc are given in Table 6.6 ofEN 1993-1-1. The additional factor is due to the critical moment being determinedelastically but failure will be by generation of plasticity not necessarily at the point ofmaximum deflection for buckling.

5.1.2.3 Simplified Assessment Methods for Building Structures(cl 6.3.2.4 EN 1993-1-1)

If there are discrete restraints to the compression flange, then lateral torsional buck-ling will not occur if the length Lc between restraints or the resultant slenderness λf

satisfies the following equation

λf = kcLc

ifzλ1= λc,0

Mc,Rd

My,Ed(5.23)

Ch05-H5060.tex 12/7/2007 15: 15 page 97


where Mc,Rd is the flexural capacity of the section, kc is the factor to correct formoment gradient, if,z is the radius of gyration of the effective compression flangewhich comprises the actual compression flange together with one-third of the partof the web which is in compression and λ1 is given by 93,9(235/fy)1/2 and λC,0 is thenormalized slenderness ratio of the effective compression flange.

5.1.2.4 Modifications Dependant upon Section Classification(cl 6.3.2.1 (3) EN 1993-1-1)

For sections other than Classes 1 and 2, a further modification is required as forClass 3 sections the capacity is based on the elastic section modulus, and Class 4 aneffective section modulus is used. The modification is made by defining the flexuralcapacity as Wel,yfy for Class 3 and Weff,yfy for Class 4. Also the buckling capacityMb,Rd is defined as χLTWpl,yfy/γM1 for Class 1 or Class 2, χLTWel,yfy/γM1 for Class 3,χLTWeff,yfy/γM1 for Class 4.

Having established the basic determination of the allowable moment capacity, Mb,Rd,two additional modifications due to varying support conditions and non-uniformflexural loading need to be examined.

5.1.3 Effect of Support Conditions

If the beam has a rotational end restraint of R (where R takes a value of 0 for norestraint and 1 for full restraint), then the effective length kL is given by Eq. (5.24)(Trahair and Bradford, 1988).

R1 − R

= − π

2kcot

π

2k(5.24)

With a little loss in accuracy Eq. (5.24) can be written as

k = 1 − 0,5R (5.25)

Both Eqs (5.24) and (5.25) are plotted in Fig. 5.3. The results from Eq. (5.25) representideal conditions therefore in practice the values of kL used are higher at more completefixity as absolute rigid joints do not exist.

Pillinger (1988) gave some indications of the relationship between practical end con-ditions and effective lengths. Figure 5.4 gives typical end conditions in terms ofconnection type and degree of restraint.

For beams the effective length L between restraints can be taken as lying between 0,7and 1,0 times the actual length with the lower factor implying full rotational restraintto both flanges and the higher factor with both flanges free to rotate in plan. If the rota-tional restraint parameter R can be assessed, then Eq. (5.25) can be modified to give

k = 1 − 0,3R (5.26)

Ch05-H5060.tex 12/7/2007 15: 15 page 98


0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

Degree or restraint (R )

Effe

ctiv

e le

ngth

fact

or (

k)

Exact solution Linear appproximation

Practical values

FIGURE 5.3 Values of effective length factor (k)

The values of kL for beams should be increased by 20% for destabilizing loads (BS5950-1: Part 1 namely 1990 and 2000).

For cantilevers the situation can be more complex as the effective length will dependon both the restraint at the encastré end and at the tip. Table 5.3 (from BS 5950-1(2000)) gives suitable values. It should be noted that when the loading is destabilizingthe effective length factors can be extremely high. Guidance is also given in Nethercotand Lawson (1992).

Although it was suggested that for beams it is generally conservative to take the actuallength, this may not be appropriate where the beam is supported solely on its bottomflange with no web or top flange restraint (Fig. 5.5). This type of situation will producelow lateral torsion buckling resistance, and Bradford (1989) suggests that in this casethe system length L should be taken as,

L = 1 + αhs

6

(twtf

)3⎛⎝1 + b

hs

2

⎞⎠ (5.27)

where the beam is under a moment gradient, or

L = 1 + 10hs

6

(twtf

)3/2⎛⎝1 + b

hs

2

⎞⎠ (5.28)

under a central point load.

Ch05-H5060.tex 12/7/2007 15: 15 page 99


Full torsionalrestraint.No rotation ofcompression flange

(g) Full end stiffener

Compressionflange rotationreduced.Torsional endrestraint

(f) Flange bolted to padstone, web stiffener

(e) Full depth end plate welded to web and flange

Full torsional endrestraint to beam

Beamcompressionflange freeto rotateon plan

Compressionflange freeto rotate.No torsionalrestraint

(c) Web cleats bolted to beam

(d) Flange bolted to padstone

Stability cleat

Fulltorsionalend restraintto beam

Beamcompressionflange freeto rotateon plan

(a) Thin end plate to web only (b) Bottom flange cleat

FIGURE 5.4 Typical connection and support detail

In both cases l is the span, hs the distance between the centroids of the flanges, b theflange width and tf and tw the thickness of the flange and web respectively, and α isgiven by

α = 4 + 7ψ + 4ψ2 (5.29)

where ψ is the ratio of the end moments.

5.1.4 Intermediate Restraints

Consider the beam in Fig. 5.6 but with a spring restraints giving a horizontal restraintstiffness of αt (force/per unit displacement) and rotational restraint stiffness of αr

Ch05-H5060.tex 12/7/2007 15: 15 page 100

TABLE 5.3 Effective length LE for cantilevers without intermediate restraint.

Restraint conditions Loading conditions

At support At tip Normal Destabilizing

(a) Continuous, with lateral (1) Free 3.0L 7.5Lrestraint to top flange (2) Lateral restraint to 2.7L 7.5L

top flange(3) Torsional restraint 2.4L 4.5L(4) Lateral and 2.1L 3.6L

torsional restraint

L

(b) Continuous, with partial (1) Free 2.0L 5.0Ltorsional restraint (2) Lateral restraint to 1.8L 5.0L


torsional restraint

L

(c) Continuous, with lateral (1) Free 1.0L 2.5Land torsional restraint (2) Lateral restraint to 0.9L 2.5L


torsional restraint

L

(d) Restrained laterally, (1) Free 0.8L 1.4Ltorsionally and against (2) Lateral restraint to 0.7L 1.4Lrotation on plan top flange

(3) Torsional restraint 0.6L 0.6L(4) Lateral and 0.5L 0.5L

torsional restraint

L

Tip restraint conditions(1) Free (2) Lateral restraint to (3) Torsional restraint (4) Lateral and torsional

top flange restraint

(not braced on plan) (braced on plan in at (not braced on plan) (braced on plan in at leastleast one bay) one bay)

Source: BS 5950 Part 1:2000

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A

A

Section AA

c of boltsL

FIGURE 5.5 Beam supported on bottom flange with no restraint to top flange

at mid-span (Fig 5.6(a)) at a height of bs above the centroidal axis. It can be shown(Mutton and Trahair, 1973) that the elastic critical moment is given by Eq. (5.6) or(5.7) with kL (the effective, or system, length) substituted for L, with k related to αt

and αr by

αtL3

16EIz

⎛⎝1 +

2bsh

2b0h

⎞⎠ =

(π

2k

)3cot π

2kπ

2k cot π

2k − 1(5.30)

αrL3

16EIw

1 − 2bsh

2b0h

=(

π

2k

)3cot π

2kπ

2k cot π

2k − 1(5.31)

where the distance of the centre of rotation below the shear centre b0 is given by

b0 = Mcr

π2 EIz(kL)2

= h2

√1 +

(kK

)2

(5.32)

and K is defined

K =(

π2EIw

L2GIt

)1/2

(5.33)

Ch05-H5060.tex 12/7/2007 15: 15 page 102


z y

bs

Axis of buckled beam

vL

φL2

2

αeφL2

α1 VL � bs φL2 2

⎛⎜⎝

⎛⎜⎝

x

FIGURE 5.6(a) Beam with elastic intermediate restraints

0 5 10 15 20 25 30

0.2

0

0.4

0.6

0.8

1

1.2

Effe

ctiv

e le

ngth

fact

or (

k)

Normalized restraint factor

Bracedmode 1 buckling

Mode 2 buckling0,5

π2

FIGURE 5.6(b) Effect of spring restraint on effective length factor

The relationships from Eqs (5.30) and (5.31) are plotted in Fig 5.6(b). The value ofk changes from 1 (when the beam buckles as an entity) to k = 0,5 when the beambuckles as two half waves when the right hand sides of Eq. (5.30) or (5.31) reach avalue of π2.

From Eq. (5.32) the minimum value of b0 is h/2. From Eq. (5.30) the maximumvalue of αt applied at the top flange (bs = h/2) to force a change into second mode isgiven by

αt = 4Mcr

Lh= 4Pf

L(5.34)

where Pf is the force in the flange.

Ch05-H5060.tex 12/7/2007 15: 15 page 103


5.1.5 Loading

The effect of the applied loading on the beam system needs considering under twoheadings. The first is concerned with how the load is applied whether at restraint pointsor between restraints. The second is concerned with any possible destabilizing effectsof the load.

5.1.5.1 Load Pattern

The background to any amendments to the value of Mcr to allow for load pattern isgiven in Nethercot and Rockey (1971), or Trahair and Bradford (1988).

For loading at points of lateral torsional restraint, the critical moment Mcr is modifiedby a factor C1 which is dependant solely on the moment gradient within the section ofthe beam being considered. The moment gradient is defined by the ratio of the appliedmoments at either end of the beam segment. The small moment within the segmentcaused by self-weight is not taken into account in this calculation.

From Kirby and Nethercot (1979) C1 is given by

1C1

= 0,57 + 0,33β1 + 0,1β21 ≥ 0,43 (5.35)

where β1 = 1,0 for single curvature (or −1 for double curvature).

And from Trahair and Bradford (1988) as either,

C1 = 1,75 + 1,05β2 + 0,3β22 ≤ 2,56 (5.36)

or,

1C1

= 0,6 − 0,4β2 ≥ 0,4 (5.37)

where β2 = −1 for single curvature (and +1 for double curvature).

Rewrite Eqs (5.35) to (5.37) using ψ rather than β1 or β2, defining ψ = 1 for singlecurvature to give

1C1

= 0,57 + 0,33ψ + 0,1ψ2 ≤ 0,43 (5.38)

C1 = 1,75 − 1,05ψ + 0,3ψ2 ≤ 2,56 (5.39)

1C1

= 0,6 + 0,4ψ ≥ 0,4 (5.40)

A comparison between the values of C1 from Eqs (5.38) to (5.39) is given in Table 5.4and Fig. 5.7 where it is observed that there is little difference between the values, andthat this difference is reduced as the value of the normalized slenderness ratio requiresthe value of C1/2

1 . For the examples herein Eq. (5.40) will be used.

Ch05-H5060.tex 12/7/2007 15: 15 page 104


TABLE 5.4 Comparison of values of C1.

ψ Eq. (5.36) Eq. (5.37) Eq. (5.38)

1,0 1,00 1,00 1,000,75 1,14 1,13 1,110,5 1,32 1,30 1,250,25 1,52 1,51 1,430 1,75 1,75 1,67

−0,25 2,03 2,03 2,00−0,5 2,33 2,35 2,50−0,75 2,33 2,56 2,50−1,0 2,33 2,56 2,50

01

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1 0

�0.1

�0.2

�0.3

�0.4

�0.5

�0.6

�0.7

�0.8

�0.9

�1

0.5

1

1.5

2

2.5

3

Moment ratio

C1

Eq. (5.36) Eq. (5.38)Eq. (5.37)

FIGURE 5.7 Values of C1

For loads between restraints use the n factor method from Tables 15 to 17 of BS 5950-1(1990) should be used. In this case C1 is defined by

C1 = 1√n

(5.41)

Tables 15 to 17 of BS 5950: Part 1: 1990 are reproduced in Annexe A7.

5.1.5.2 Destabilizing and Stabilizing Loads

A destabilizing load is one which is applied to the compression flange and is freeto move as the flange buckles laterally. Such a load has the effect of reducing theelastic critical moment as an additional disturbing torque is introduced (Andersonand Trahair, 1972; Trahair and Bradford, 1988). A stabilizing load, therefore, is one

Ch05-H5060.tex 12/7/2007 15: 15 page 105


Additionaltorque

No additionaltorque

Decreasedtorque

FIGURE 5.8 Destabilizing loads

applied in such a way that the effect counterbalances the buckling effect and thusenhances the elastic critical moment. Thus for simply supported beams if a load freeto move is applied above the shear centre it is destablizing, and below the shear centreit is stabilizing (Fig. 5.8). The reverse is true for a cantilever!

For loading applied at other than the centroid of the section it is only possible todetermine the critical moment under a given load pattern.

Timoshenko and Gere (1961) give the value of the critical load for a simply supportedI beam under either a uniformly distributed load (UDL) or central point load as

Pcr = γ2

√EIzGIt

L2 (5.42)

where γ2 is dependant upon the position of the load (top flange, centroid or bottomflange) and the factor L2GI t/EIw. Values of γ2 are plotted in Fig. 5.9 for values ofbeam stiffness 1/K2 (where K is given by Eq. (5.33)).

5.1.5.3 Cantilevers and Beams Cantilevering Over Supports

• Cantilevers

Trahair (1983) demonstrated that for cantilevers built in at the support that the elasticcritical moment Mcr could be given with little loss of accuracy as

Mcr =√

EIzGIt

L(1,6 + 0,8K) (5.43)

where K is given by Eq. (5.33) as√

(π2EIw/(L2GI t)).

It will be noted that this differs from the more usual solution given in Eq. (5.6) wherethe elastic critical moment is proportional to (1 + K2)1/2.

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0

20

40

60

80

100

120

140

160

0 50 100 150 200 250 300 350 400 450

Beam stiffness (1/K 2)

Gam

ma

2

Load at top Load at centre Load at bottom

FIGURE 5.9(a) Factor for point load not applied at shear centre

0

50

100

200

150

250

0 50 100 150 200 250 300 350 400 450

Beam stiffness (1/K 2)

Gam

ma

2

Load at top Load at centre Load at bottom

FIGURE 5.9(b) Factor for UDL not applied at shear centre

For a point load Pcr at the free end of the cantilever Trahair gives the followingexpression,

Pcr =√

EIzGIt

L2

[11

(1 + 1,2ε√

1 + 1,22ε2

)+ 4(K − 2)

(1 + 1,2(ε − 0,1)√

1 + 1,22(ε − 0,1)2

)]

(5.44)

Ch05-H5060.tex 12/7/2007 15: 15 page 107


where the parameter ε defines the position relative to the centroidal axis of the pointof application of the load, and is given by

ε = aL

√EIz

GIt= 2a

hKπ

(5.45)

where a is the distance of the point of application of the load

For a UDL of qcr applied to whole length of the cantilever,

qcr = 2√

EIzGIt

L3

[27

(1 + 1,4 (ε − 0,1)√

1 + 1,42(ε − 0,1)2

)

+ 10(K − 2)

(1 + 1,3(ε − 0,1)√

1 + 1,32(ε − 0,1)2

)](5.46)

For loading applied at the centroid, ε = 0, and Eq. (5.44) reduces to

Pcr =√

EIzGIt

L2 [3,96 + 3,52K ] (5.47)

The moment Mcr,P due to this load is given by

Mcr,P = PcrL =√

EIzGIt

L[3,96 + 3,52K ] (5.48)

Equation (5.46) becomes when loading is applied at the centroid

qcr = 2√

EIzGIt

L3 [5,83 + 8,71K ] (5.49)

The moment Mcr,q due to this load is given by

Mcr,q = qcrL2

2=

√EIzGIt

L[5,83 + 8,71K ] (5.50)

Equations (5.43), (5.48) and (5.50) are plotted in Fig. 5.10, where it is observed thatas the loading progresses from uniform moment through a point load at the end toa UDL, the equivalent elastic moment increases. This is due to the fact that for theUDL the moment quickly drops off from the support and thus has less of an effect.The moment due to the point load also drops off, but more slowly. It is therefore tobe noted that the general UK practice of ignoring the beneficial effects of momentgradients on a cantilever (i.e. C1 = 1,0) is extremely conservative.

• Overhanging beams

For the situation shown in Fig. 5.11(a), where a beam overhangs the supports sym-metrically and is loaded by end point loads, then the lateral torsional buckling ofsuch a system is controlled by the restraint at the internal supports which mayoften be of the type shown in Fig. 5.11(b), where such restraints can be consid-ered elastic. Such restraints increase the torsional flexibility of the beam and therebydecrease the buckling moment. The decrease P/P0 is dependant upon the stiffness

Ch05-H5060.tex 12/7/2007 15: 15 page 108


0

5

10

15

20

25

30

35

0 0.5 1 1.5 2 2.5 3 3.5

Stiffness parameter K

Nor

mal

ized

ela

stic

crit

ical

mom

ent

Point loadPure moment UDL

FIGURE 5.10 Normalized elastic critical moment for a cantilever

(a) Basic geometry

(b) Type support detail

P P

γLL L

Over hanging beam

Supporting beam

Bolts

FIGURE 5.11 Overhanging beams

at the support αR, and Trahair (1983) gives the following equation for beams loadedcentroidally,

PP0

=(

1 − 2ah

Kβ2 (1 − β)

)√√√√√ αRLGIt

5 + 4K2

1+K2 + αRLGIt

(5.51)

Ch05-H5060.tex 12/7/2007 15: 15 page 109


40 kN Permanent70 kN Variable

20 kN Permanent30 kN Variable

A D

CB

3.0 m 3.0 m 3.0 m

Notes: (1) All loads are characteristic loads (2) Lateral torsional restraints exists at A, B, C and D

FIGURE 5.12 Design data for Example 5.1

where β is defined by

β =αRLGIt

1 + αRLGIt

(5.52)

It should be noted for situations where the internal span is greater than the sum of theoverhangs, then the internal span dominates the behaviour.

EXAMPLE 5.1 Beam with loading applied at restraints.

Prepare a design in Grade S355 steel for the beam for which the data are given inFig. 5.12.

Factored actions at ULS:

at B: 1,35 × 40 + 1,5 × 70 = 159 kNat C: 1,35 × 20 + 1,5 × 30 = 72 kN

Total load = 231 kN

The BM and SF diagrams are drawn in Fig. 5.13.

The critical section for design is the central section BC as the moment gradient is theleast.

Try a 406 × 178 × 74 UKB

Mpl,Rd = Wpl,yfy

γM0= 1501000

3551,0

× 10−6 = 533 kNm

MSd = 390 kNm, beam satisfies the plastic capacity criterion.

Section classification:

Compression flange:

c = 0,5[b − 2r − tw] = 0,5[179,5 − 2 × 10,2 − 9,5] = 74,8 mm

c/tf = 74,8/16 = 4,68

Ch05-H5060.tex 12/7/2007 15: 15 page 110


A

B C D

303

390

A B C D

(a) BMD (kNm)

(b) SFD (kN)

130

−29

−101FIGURE 5.13 BMDand SFD forExample 5.1

Maximum value for a Class 1 flange is 9ε = 9(235/355)1/2 = 7,32.

Web:

c/t = d/tw = 37,9.

Maximum value for a Class 1 web is 9ε = 72(235/355)1/2 = 58,6.

Thus a 406 × 178 × 74 UKB Grade S355 is Class 1.

Shear check:

Vpl,Rd = 1√3

Avfy

γM0= 1√

34260

3551,0

× 10−3 = 873 kN

By inspection, the moment capacity is not reduced due to shear.

Calculation of Mcr:

E = 210 GPa; G = 81 GPa.

The most foolproof method to determine Mcr given the use of the cm and dm in sectionproperty tables is to work in kN and m.

Use Eq. (5.5) to determine Mcr.

System length L = 3 m.

Iw

Iz= 0,608 × 10−6

1545 × 10−8 = 0,03924 m2

π2EIz

L2 = π2 × 210 × 106 × 1545 × 10−8

32 = 3558 kN

Ch05-H5060.tex 12/7/2007 15: 15 page 111


L2GIt

π2EIz= GIt

π2EIzL2

= 81 × 106 × 62,8 × 10−8

3558= 0,0143 m2

Mcr = π2EIz

L2

[Iw

Iz+ L2GIt

π2EIz

]1/2

= 3558 [0,03924 + 0,0143]1/2 = 823 kNm

Determine the value of C1 using Eq. (5.40):

The moment ratio ψ is given by

ψ = MEd,C

MEd,B= 303

390= 0,777

1C1

= 0,6 + 0,4ψ = 0,6 + 0,4 × 0,777 = 0,911

or

C1 = 10,911

= 1,098

The value of Mcr to be used in Eq. (5.12) in the calculation of the normalizedslenderness ratio is C1Mcr, thus λLT is given as

λLT =√

WyfyMcr

=√

1501000 × 3551,098 × 823 × 106 = 0,768

Both methods available to determine the strength reduction factors due to lateraltorsional buckling will be used in this example to demonstrate any differences.

(a) General case

The ratio h/b = 412,8/179,5 = 2,3 > 2, so from Table 5.1, αLT = 0,34 (curve b).

Use Eqs (5.15) and (5.16) to determine �LT:

�LT = 0,5[1 + ηLT + (λLT)2] = 0,5[1 + 0,34(0,768 − 0,2) + 0,7682] = 0,891

Use Eq. (5.14) to determine χLT:

χLT = 1�LT + [�2

LT − (λLT)2]1/2= 1

0,891 + [0,8912 − 0,7682]1/2 = 0,745

Mb,Rd = χLTWpl,yfy

γM1= 0,745 × 1501000

3551,0

× 10−6 = 397 kNm

This is greater than the moment at B (390 kNm).

Clearly the end section AB does not need checking as the system length is the same asBC (therefore the basic value of Mcr does not change and since ψ = 0, C1 now becomes1,75 with the effect of reducing the value of λLT and of increasing the value of χLT andhence Mb,Rd.

(b) Method for rolled sections

The ratio h/b = 412,8/179,5 = 2,3 > 2, so from Table 5.2, αLT = 0,49 (curve c).

Ch05-H5060.tex 12/7/2007 15: 15 page 112


Determine ηLT from Eq. (5.19)

ηLT = αLT(λLT − λLT,0) = 0,49(0,768 − 0,4) = 0,180

Determine �LT from Eq. (5.18)

�LT = 0,5[1 + ηLT + β(λLT)2] = 0,5[1 + 0,180 + 0,75 × 0,7682] = 0,811

Determine χLT from Eq. (5.17)

χLT = 1�LT + [�2

LT − β(λLT)2]1/2= 1

0,811 + [0,8112 − 0,75 × 0,7682]= 0,784

Initially ignore the correction factor f , to give Mb,Rd as

Mb,Rd = χLTWpl,yfy

γM1= 0,784 × 1501000

3551,0

× 10−6 = 418 kNm

Determine f :

The moment ratio ψ = 303/390 = 0,777, so from Eq. (5.33),

kc = 11,33 − 0,33ψ

= 11,33 − 0,33 × 0,777

= 0,931

From Eq. (5.32) f is given by

f = 1 − 0,5(1 − kc)[1 − 2,0(λLT − 0,8)2] ≤ 1,0

= 1 − 0,5(1 − 0,931)[1 − 2,0(0,768 − 0,8)2] = 0,966

Using Eq. (5.31) Mb,Rd now becomes 418/0,966 = 433 kNm. The effect of the factor fis not terribly significant.

Even without the factor, f lateral torsional buckling clauses for rolled sectionsproduces a marginally higher value of Mb,Rd by around 20 kNm (or around 5%).

Deflection (under service loading):

EI = 210 × (27430 × 104) × 10−6 = 57603 kNm2

Mid-span deflection due to an asymmetric point load is given by

δ = WL3

48EI

(3(

bL

)− 4

(bL

)3)

Variable action deflection:

Load at B: W = 70 kN, b = 3 m, l = 9 m, δ = 0,016 m

Load at C: W = 30 kN, b = 3 m, L = 9 m, δ = 0,007 m

Deflection under service variable actions = 0,023 m.

This deflection is equivalent to span/390, and is therefore acceptable.

Ch05-H5060.tex 12/7/2007 15: 15 page 113


Deflection under total actions:

Load at B: W = 110 kN, b = 3 m, l = 9 m, δ = 0,025 m

Load at C: W = 50 kN, b = 3 m, L = 9 m, δ = 0,011 m

Total deflection = 0,036 m.

This deflection is equivalent to span/250, and is therefore acceptable.

Web check (refer to Section 4.8 and cl 6 EN 1993-1-5):

The web capacity under transverse forces needs checking at A and B.

Note as a rolled section is being used, fyw = fyf = 355 MPa. Since at either point thelength of stiff bearing ss is not known, set ss = 0 and determine the value requiredshould the check fail

At A: FSd = RA = 130 kN.

For an end support with c = ss = 0, kF = 2 (type c)

Determine m1:

m1 = fyfbf

fywtw= bf

tw= 179,5

9,5= 18,9

As m2 is dependant upon λF initially assume m2 = 0.

As ss and c have been assumed to be zero, then lc = 0, then the least value of lyis given by

ly = tf

√m1

2= 16,0

√18,9

2= 49,2 mm

The depth of the web hw has been taken as d, the depth between fillets.

FCR = 0,9kFEt3w

hw= 0,9 × 2 × 210

9,53

360,4= 889 kN

λF =√

lytwfyw

FCR=√

49,2 × 9,5 × 355889 × 103 = 0,432

As λF < 0.5, m2 = 0.

χF = 0,5λF

= 0,50,432

= 1,16

The maximum value of χF is 1,0, thus

Leff = χFly = 1,0 × 49,2 = 49,2 mm

FRd = Leff twfyw

γM1= 49,2 × 9,5

3551,0

× 10−3 = 166 kN

As FRd is greater than FEd (=RA = 130 kN), η2 < 1,0, therefore the web resistance atA is satisfactory without a stiff bearing.

Ch05-H5060.tex 12/7/2007 15: 15 page 114


Check at B:

The applied load is 159 kN, and the applied moment is 390 kNm.

For the situation where the load is applied through the top flange, kF = 6 (type a, withthe stiffener spacing a effectively taken as infinity)

Determine m1:

m1 = fyfbf

fywtw= bf

tw= 179,5

9,5= 18,9


As ss has been assumed to be zero, then the value of ly is given by

ly = 2tf (1 + √m1) = 2 × 16,0(1 +√18,9) = 171 mm


FCR = 0,9kFEt3w

hw= 0,9 × 6 × 210

9,53

360,4= 2698 kN

λF =√

lytwfyw

FCR=√

171 × 9,5 × 3552698 × 103 = 0,462

As λF < 0,5, m2 = 0.

χF = 0,5λF

= 0,50,462

= 1,08


Leff = χFly = 1,0 × 171 = 171 mm

FRd = Leff twfyw

γM1= 171 × 9,5

3551,0

× 10−3 = 577 kN

η2 = FEd

Leff twfywγM1

= FEd

FRd= 159

577= 0,276

η2 ≤ 1,0, therefore the web resistance at A is satisfactory without a stiff bearing.

However an interaction equation needs checking owing to the co-existence of shearand bending moment:

η2 + 0,8η1 ≤ 1,4

As there is no axial force and no shift in the neutral axis as the section is Class 1, theequation for η1 reduces to

η1 = MEd

Wplfy

γM1

= 390 × 106

1501 × 103 3551,0

= 0,732

η2 + 0,8η1 = 0,276 + 0,8 × 0,732 = 0,862 ≤ 1,4

The web at B is therefore satisfactory.

Ch05-H5060.tex 12/7/2007 15: 15 page 115


EXAMPLE 5.2 Beam design with loads applied between lateral torsional restraints.

Prepare a design in Grade S355 steel for the beam whose data are given in Fig. 5.14

Factored actions:

Factored UDL: 1,35 × 10 + 1,5 × 20 = 43,5 kN/m

Factored point load: 1,5 × 15 = 22,5 kN

Figures 5.15 (b) and (c) show the resultant BM and SF diagrams.

Try a 457 × 191 × 98 UKB Grade S355


Flanges:

c = 0,5(b − 2r − tw) = 0,5(192,8 − 2 × 10,2 − 11,4) = 80,5 mm

ctf

= 80,519,6

= 4,11

Class 1 limit:

9ε = 9

√235355

= 7,32

Flanges are Class 1.

Web:

ctw

= dtw

= 407,611,4

= 35,8

Class 1 limit:

72ε = 72

√235355

= 58,56

Web is Class 1, therefore the section classification is Class 1.

A B

C

20 kN/m Variable10 kN/m Permanent

15 kN Variable

7.0 m 3.5 m

Notes: (1) All loading is characteristic loading (2) Lateral torsional restraints exist at A and B (3) The TOP flange is restrained at C

FIGURE 5.14 Design data for Example 5.2

Ch05-H5060.tex 12/7/2007 15: 15 page 116


A B

C

22,5 kN43,5 kN/m

(a) Beam loading at ULS

(b) SFD (kN)

(c) BMD (kNm)

2,37174,7

22,5

−201,6

102,9

D BC

B

C

A D

−192,3

345,2

FIGURE 5.15 BMD and SFD for Example 5.2

Plastic moment capacity:

Mpl,Rd = Wpl,yfy

γM0= 2232 × 103 355

1,0× 10−6 = 793 kNm

This exceeds the maximum applied moment of 345,2 kNm.

Shear capacity:

Vpl,Rd = 1√3

Avfy

γM0= 1√

35590

3551,0

× 10−3 = 1146 kN

Ch05-H5060.tex 12/7/2007 15: 15 page 117


By inspection, there is no reduction in moment capacity for the effect of shear at thecantilever support.

System length AB:

The system length is taken as the span AB, not the distance between the points ofcontraflexure calculated from the applied loading as these will not form nodal pointsin the post-buckled shape of the beam. Such nodal points may only form at points oflateral restraint since at a nodal point the lateral deflection must be zero (Kirby andNethercot, 1979).

Determination of C1:

As the loading is between restraints at A and C, the ‘n’ factor method from BS 5950:Part 1: 1990 must be used (Annexe A7).

From Table 17, the BMD is that of the fifth diagram with β = 0 (as the BM at A iszero), and γ is negative.

The bending moment M0 assuming the beam to be simply supported between A andC is given by

M0 = qultL2

8= 43,5 × 72

8= 266,4 kNm

γ = MM0

= −345,2266,4

= −1,3

From Table 16 with β = 0 and γ = −1,3, n = 0,53

C1 = 1√n

= 1√0,53

= 1,374

Determine Mcr from Eq. (5.5):

Iw

Iz= 1,18 × 10−6

2347 × 10−8 = 0,0503 m2

L = 7,0 m.

π2EIz

L2 = π2 × 210 × 106 × 2347 × 10−8

72 = 993 kN

L2GIt

π2EIz= GIt

π2EIzL2

= 81 × 106 × 121 × 10−8

993= 0,0987 m2

Mcr = π2EIz

L2

[Iw

Iz+ L2GIt

π2EIz

]1/2

= 993[0,0503 + 0,0987]1/2 = 383 kNm

Determine λLT using the moment gradient modified value of Mcr in Eq. (5.12),

λLT =(

Wpl,yfyC1Mcr

)1/2

=(

2232 × 103 × 355 × 10−6

1,374 × 383

)1/2

= 1,227

Ch05-H5060.tex 12/7/2007 15: 15 page 118


As in the previous example calculate the strength reduction factor using both methods:

• General method

h/b = 358,0/172,2 = 2,08 > 2, thus from Table 5.1, αLT = 0,34

Determine �LT from Eqs (5.15) and (5.16)

�LT = 0,5[1 + αLT(λLT − 0,2) + (λLT)2]

= 0,5[1 + 0,34(1,227 − 0,2) + 1,2272] = 1,427


χLT = 1

�LT + [�2LT − (λLT)2

]1/2 = 11,427 + [1,4272 − 1,2272]1/2 = 0,464

Mb,Rd = χLTWpl,yfy

γM1= 0,464 × 2232 × 103 355

1,0× 10−6 = 368 kNm

This exceeds the absolute value of the maximum moment at C of 345,2 kNm.

• Rolled section method:



�LT = 0,5[1 + αLT(λLT − λLT,0) + β(λLT)2]

= 0,5[1 + 0,49(1,227 − 0,4) + 0,75 × 1,2272] = 1,267


χLT = 1�LT + [�2

LT − β(λLT)2]1/2= 1

1,267 + [1,2672 − 0,75 × 1,2272]1/2 = 0,511

Without the correction factor f :

Mb,Rd = χLTWpl,yfy

γM1= 0,511 × 2232 × 103 355

1,0× 10−6 = 405 kNm


From Table 6.6 of EN 1993-1-1, kc = 0,91

Determine f from Eq. (5.32):

f = 1 − 0,5(1 − kc)[1 − 2,0(λLT − 0,8)2]

= 1 − 0,5(1 − 0,91)[1 − 2,0(1,227 − 0,8)2] = 0,971

Thus the corrected value of Mb,Rd given by Eq. (5.31) is

Mb,Rd = 4050,971

= 417 kNm

It is again noted that the rolled section approach even without the factor f gives slightlyhigher values of Mb,Rd than the general case.

Ch05-H5060.tex 12/7/2007 15: 15 page 119


System length BC.

Take the conventional approach and adopt C1 = 1,0

Determine Mcr from Eq. (5.5):

Iw

Iz= 1,18 × 10−6

2347 × 10−8 = 0,0503 m2

L = 3,5 m

π2EIz

L2 = π2 × 210 × 106 × 2347 × 10−8

3,52 = 3971 kN

L2GIt

π2EIz= GIt

π2EIzL2

= 81 × 106 × 121 × 10−8

3971= 0,0247 m2

Mcr = π2EIz

L2

[Iw

Iz+ L2GIt

π2EIz

]1/2

= 3971[0,0503 + 0,0247]1/2 = 1088 kNm

Determine λLT from Eq. (5.12),

λLT =(

Wpl,yfyC1Mcr

)1/2

=(

2232 × 103 × 355 × 10−6

1088

)1/2

= 0,853

As in the previous example calculate the strength reduction factor using both methods:

• General method



�LT = 0,5[1 + αLT(λLT − 0,2) + (λLT)2]

= 0,5[1 + 0,34(0,853 − 0,2) + 0,8532] = 0,975


χLT = 1

�LT + [�2LT − (λLT)2

]1/2 = 10,975 + [0,9752 − 0,8532]1/2 = 0,691

Mb,Rd = χLTWpl,yfy

γM1= 0,691 × 2232 × 103 355

1,0× 10−6 = 548 kNm


• Rolled section method



�LT = 0,5[1 + αLT(λLT − λLT,0) + β(λLT)2]

= 0,5[1 + 0,49(0,853 − 0,4) + 0,75 × 0,8532] = 0,884

Ch05-H5060.tex 12/7/2007 15: 15 page 120


A B

P

CEl constant

a b

Deflection: At mid-span (span AB) At C

Deflection: At mid-span (span AB) At C

A B C

El constant

a b

(qa2/384 EI )(5a2 − 12b2)(qb/24 EI)(3b3 + 4ab2 − a3)

−Pba2/16EIPb2(a + b)/3EI

(a) UDL

(b) Point load

FIGURE 5.16 Deflection formulae for Example 5.2


χLT = 1�LT + [�2

LT − β(λLT)2]1/2

= 10,884 + [0,8842 − 0,75 × 0,8532]1/2 = 0,730

As there appears to be no consideration given to cantilevers in Table 6.6, f will betaken as 1,0.

Mb,Rd = χLTWpl,yfy

γM1= 0,730 × 2232 × 103 355

1,0× 10−6 = 578 kNm


It is again noted that the rolled section approach even without the factor f gives slightlyhigher values of Mb,Rd than the general case.

For this particular design case, the span AC is critical.

Deflection check:

EI = 210 × 10−6 × 45730 × 104 = 96033 kNm2.

The relevant formulae are given in Fig. 5.16.

(a) Variable action check:UDL: 20 kN/m

Ch05-H5060.tex 12/7/2007 15: 15 page 121


Span AB, central deflection

δ = qa2

384EI(5a2 − 12b2) = 20 × 72

384 × 96033(5 × 72 − 12 × 3,52) = 0,0026 m

Span BC, at C

δ = qb24EI

(3b3 + 4ab2 − a3) = 20 × 3,524 × 96033

(3 × 3,53 + 4 × 7 × 3,52 − 73)

= 0,0039 m

Point load at CSpan AB, mid-span

δ = − Pba2

16EI= −15 × 3,5 × 72

16 × 96033= −0,0017 m

Span BC, at C

δ = P(a + b)b2

3EI= 15(7 + 3,5)3,52

3 × 96033= 0,0067 m

Net deflections:at mid-span = 0,0026 − 0,0017 = 0,0009 mSpan deflection ratio is 7/0,0009 = 7780. This is more than acceptable.at C = 0,0039 + 0,0067 = 0,0106 mSpan deflection ratio (based on twice the span) is 2 × 3,5/0,0106 = 660.This is acceptable.

(b) Check under total actions.Deflection due to point load as above.Total UDL of 30 kN/m:Mid-span, δ = 0,0039 m; at C, δ = 0,0059 m.Total deflection at mid-span = 0,0039 − 0,0017 = 0,0022 mSpan deflection ratio: 7/0,0022 = 3180. This is satisfactory.Total deflection at C = 0,0059 + 0,0067 = 0,0126 mSpan deflection ratio (based on twice the span) is 2 × 3,5/0,0126 = 556.This is satisfactory.

Web check at B (refer to Section 4.8 and cl 6 EN 1993-1-5):

This is the only point that needs checking, as the other reaction point has a much lowerforce (even allowing for reduced dispersion length) and no coincident moment.

RSd = 124,7 kN (Reaction at B); M = 114,5 kNm

Ignore any stiff bearing (ss = 0).

For the situation where the load is applied through the top flange, kF = 6 (type a, withthe stiffener spacing a effectively taken as infinity)

Ch05-H5060.tex 12/7/2007 15: 15 page 122


Determine m1:

m1 = fyfbf

fywtw= bf

tw= 192,8

11,4= 16,9


As ss has been assumed to be zero, then the value of ly is given by

ly = 2tf (1 + √m1) = 2 × 19,6(1 +√16,9) = 200 mm


FCR = 0,9kFEt3w

hw= 0,9 × 6 × 210

11,43

407,6= 4122 kN

λF =√

lytwfyw

FCR=√

200 × 11,4 × 3554122 × 103 = 0,443

As λF < 0,5, m2 = 0.

χF = 0.5λF

= 0,50,443

= 1,13


Leff = χFly = 1,0 × 200 = 200 mm

FRd = Leff twfyw

γM1= 200 × 11,4

3551,0

× 10−3 = 809 kN

η2 = FEd

Leff twfyw

γM1

= FEd

FRd= 124,7

809= 0,154

η2 ≤ 1,0, therefore the web resistance at A is satisfactory without a stiff bearing.

However an interaction equation needs checking owing to the co-existence of shearand bending moment:

η2 + 0,8η1 ≤ 1,4


η1 = MEd

Wplfy

γM1

= 114,5 × 106

2232 × 103 3551,0

= 0,145

η2 + 0,8η1 = 0,154 + 0,8 × 0,145 = 0,27 ≤ 1,4


Ch05-H5060.tex 12/7/2007 15: 15 page 123


5.1.6 Other Section Profiles

A number of special cases need considering, hollow sections as the earlier equationsfor critical moment are not applicable, rectangular sections as the warping stiffness iszero, and T sections.

5.1.6.1 Rolled Hollow Sections

Rees (1990) indicates that thin wall tubes of circular and triangular cross with uni-form thickness cannot warp. For rectangular thin walled tubes only those whose wallthicknesses are in a constant ratio the length of the sides do not warp. These tubesare known as Neuber tubes. If there is no warping then lateral torsional buckling canonly be resisted by torsion. For conventional hollow sections where the wall thicknessis constant, then lateral torsional buckling is in part resisted by warping. It will beconservative to neglect the warping stiffness, and, therefore as a result, Eq. (5.3) canbe reduced to

GItd2φ

dx2 = −M2

EIzφ (5.53)

The resultant value of Mcr (with no allowance for major axis bending) is given fromEq. (5.6) as

Mcr = π√

EIzGIt

L(5.54)

The factor K from Eq. (5.7) with the warping constant Iw set equal to zero becomes,

K =√(

1 − EIz

EIy

)(1 − GIt

EIy

)(5.55)

Combining Eqs (5.54) and (5.55) gives the elastic critical moment as

Mcr = π√

EIzGIt

L√(

1 − EIzEIy

) (1 − GIt

EIy

) (5.56)

or the normalized slenderness ratio, λLT, is given from Eq. (5.12) with the introductionof the moment gradient factor C1 as

λLT =√

Wpl,yfyC1Mcr

=

√√√√√Wpl,yfyL√(

1 − IzIy

) (1 − GIt

EIy

)C1π

√EIzGIt

(5.57)

The normalized lateral torsional buckling slenderness ratio λLT is also given by

λLT = λLT

λ1(5.58)

Ch05-H5060.tex 12/7/2007 15: 15 page 124


or

λLT = λLTλ1 = π

√Efy

λLT =

√√√√√πEWpl,yL√(

1 − IzIy

) (1 − GIt

EIy

)C1

√EIzGIt

(5.59)

Rewrite Eq. (5.59) as

λLT = 1

C1/21

[π

√EG

]1/2

√√√√√Wpl,yL√(

1 − IzIy

) (1 − GIt

EIy

)√

IzIt(5.60)

Define the slenderness ratio λ as

λ = Liz

= L√IzA

(5.61)

when Eq. (5.62) becomes

λLT = 1

C1/21

[π

√EG

]1/2

√√√√√Wpl,yλ

√(1 − Iz

Iy

) (1 − GIt

EIy

)√

AIt(5.62)

or λLT is given as

λLT = 1

C1/21

[π

√EG

]1/2

(φbλ)1/2 (5.63)

where φb is defined as by

φb =⎛⎝W 2

pl,y

[1 − Iz

Iy

] [1 − GIt

EIy

]AIt

⎞⎠

1/2

(5.64)

This is the equation given in Section B2.6.1 of BS 5950-1. The term in rectangularparentheses in Eq. (5.63) has a value of 2,25.

An alternative approach avoiding the calculation of lateral torsion buckling is to deter-mine the critical length lcrit (in mm) corresponding to a value of λLT equalling 0,4(below which buckling will not occur) (Rondal, et al., 1992). This value is given by

lcrit = 113400(h − t)fy

(b−th−t

)2

1 + 3 b−th−t

√√√√3 + b−th−t

1 + b−th−t

(5.64a)

It should be noted that there is an apparent anomaly in Eq. (5.64) in that for a squaresection (b = h), lcrit remains finite, whereas Eq. (5.57) indicates φb (and hence λLT) = 0.

Ch05-H5060.tex 12/7/2007 15: 15 page 125


The reason is that the values from Eq. (5.64) in Rondal, et al. are also given in tabularform for discrete values of (h − t)/(b − t), and that interpolation would not be possibleif the value for a square section were given as infinity. It should be noted that there isan error in the formula quoted in Rondal, et al., but that the values in Table 15 in thesame publication are correct. Eq. (5.168) has been corrected (Kaim, 2006). It shouldbe noted that the values of lcrit are extremely safe.

Kaim (2006) suggests the normalized slenderness limit λz,lim is given by

λz,lim = 25hb

√235fy

(5.64b)

5.1.6.2 Rectangular Sections

For rectangular sections of width b and depth h the equation for critical moment Mcr

given in Eq. (5.6) reduces to

Mcr = π

L

√EIzGIt (5.65)

as the warping constant Iw is zero.

(a) Thin sections

For thin sections, It = hb3/3 and Iz = hb3/12, so Eq. (5.65) reduces to

Mcr = hb3

LπE√

72(1 + ν)(5.66)

(b) Thick sections

In this case It is no longer given by ht3/3. The following approximate formula canbe used

It = hb3

3

[1 − 0,63

bh

(1 − 1

12

(bh

)4)]

(5.67)

If h/b > 2, Eq. (5.67) can with little loss in accuracy be reduced to

It = hb3

3

[1 − 0,63

bh

](5.68)

However the major axis bending is now important and to ignore it would be tooconservative, thus the parameter K from Eq. (5.8) must be introduced to give Mcr as

Mcr = πhb3

6L

√EG

√√√√√1 − 0,63 bh

1 −(

bh

)2 (5.69)

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Axis throughshear centre

Elasticneutral axisd

t 1t 2

b1

b2

tw

h s

z c

z 0

FIGURE 5.17Calculation ofmonosymmetry index

5.1.6.3 Monosymmetric Beams

For beams with only one axis of symmetry (usually the minor axis), the centroid andthe shear centre do not coincide, thus an additional disturbing torque occurs due tothe longitudinal flexural stresses.

From Trahair and Bradford (1988) the elastic critical moment Mcr for the monosym-metric beam in Fig. 5.17 is given by

Mcr = π

L

√EIzGIt

⎧⎨⎩√[

1 + π2EIw

GItL2 +(πγM

2

)2]

+ πγM

2

⎫⎬⎭ (5.70)

where γM is given by

γM = βy

L

√EIz

GIt(5.71)

The monosymmetry parameter for the section βy is given by

βy = 1Iy

{βy1 − βy2 + βy3

}− 2z0 (5.72)

where

βy1 = (hs − zc)

[b3

2t212

+ b2t2(hs − zc)

](5.73)

βy2 = zc

[b3

1t112

+ b1t1z2c

](5.74)

Ch05-H5060.tex 12/7/2007 15: 15 page 127


βy3 = tw4

[(hs − zc − t2

2

)4

−(

zc − t12

)4]

(5.75)

where

hs = d − t1 + t22

(5.76)

zc = b2t2hs + tw2 (d − t1 − t2)(d − t2)

b1t1 + b2t2 + (d − t1 − t2)tw(5.77)

z0 = αhs − zc (5.78)

α = 1

1 +(

b1b2

)3 t1t2

(5.79)

The warping constant Iw is now given by

Iw = α(1 − α)Izh2s (5.80)

5.1.6.4 ‘T’ Beams

For ‘T’ beams, it should be first checked that the value of K from Eq. (5.8) is real asa large number of commercial ‘T’ beams have Iy > Iz in which case lateral torsionalbuckling cannot occur. Where Mcr needs calculating, it should be noted Iw is zero asα = 0 (Eq. (5.80)) and z0 = −zc. The position of the centroid zc and Iy are tabulated insection property tables.

5.1.6.5 Parallel Flange Channels

The procedure for calculating Mcr follows that for ‘I’ beams except that Iw iscalculated as

Iw = tfb3f h2

123bf tf + 2htw6bf tf + htw

(5.81)

where bf and tf are the width and thickness of the flange and h and tw are the heightand thickness of the web. For channels with tapered flanges tf may be taken as themean thickness of the flange (Kirby and Nethercot, 1979).

5.2 PURE TORSIONAL BUCKLING

This form of failure can only occur in open sections and is most likely to only wherethe sections are thin walled. There can then be an interaction between strut bucklingor buckling in pure torsion.

Ch05-H5060.tex 12/7/2007 15: 15 page 128


x

x

σ

Thickness (t )

b

b

b b

dθdd

scL

φ

dv

dH

φ

(b) Cross-section(a) Basic layout

(c) Deflected geometry FIGURE 5.18 Cruciform strut

5.2.1 Interaction between Torsional Buckling and Strut Buckling

This is best illustrated by considering the case of a cruciform strut loaded by a uniformstress σ over its cross-section. To demonstrate the principles a thin section strut isconsidered (Fig. 5.18). The load induced by the stress σ remains parallel to the x axis,and therefore induces a lateral force in the yz plane. If the vertical component inFig 5.18(c) is dV then the horizontal component dH is given by

+dH = dV tan φ = φdV = sdθ

dxdV (5.82)

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as φ is small, and where dθ/dx is the angle of twist per unit length and s is a distancemeasured from the x axis.

The force dV is given by

dV = σdA = σtds (5.83)

where t is the thickness. The incremental torque dT is given by

dT = sdH (5.84)

Substitute Eqs (5.81) and (5.83) into Eq. (5.84) to give

dT = σdθ

dxts2ds (5.85)

Integrate Eq. (5.85) over the four arms of the strut to give

T = 4σdθ

dxt∫ h

0s2ds = σ

dθ

dx43

b3t (5.86)

where 2b is the width of the strut.

Equation (5.86) may be rewritten as

T = σdθ

dxIx (5.87)

where Ix is the polar second moment of area about the x axis, and is given by

Ix = 43

b3t (5.88)

The second moment of area of the strut about either the z or y axis is given by

Iz = Iy = 112

(2b)3t = 23

b3t (5.89)

as the other arm of the strut has negligible second moment of area (bt3/6) comparedwith the other direction.

Note that

Ix = Iy + Iz = 2(

23

b3t)

= 43

b3t (5.90)

From torsion theory the torque that may be carried by the section is given by

T = GItdθ

dx(5.91)

where G is the shear modulus and Tt is the torsional second moment of area. For thecruciform thin walled section It is given by

It = 4(

13

bt3)

= 43

bt3 (5.92)

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Strutbuckling

6s0

π2 E

Lb

bt

Yield

Torsionalbuckling

2(1�v)σ0

E

FIGURE 5.19 Interaction diagram

Substitute Eq. (5.92) into Eq. (5.91) to give

T = 43

bt3Gdθ

dx(5.93)

For the section to be able to sustain its twisted shape the two values of the torquesfrom Eqs (5.87) and (5.93) must be equal, so

σ = G(

tb

)2

(5.94)

There also exists the possibility that the strut can undergo normal Euler buckling undera stress σcr given by

σcr = π2EIz

AL2 = π2E 23 b3t

4btL2 = π2E6

(bL

)2

(5.95)

Assuming no interaction, there are three possibilities of behaviour; the stress producesyield; torsional buckling occurs, or strut buckling occurs (Fig. 5.19). The stress whichcauses torsional buckling and strut buckling to occur simultaneously is given when thestresses from Eqs (5.94) and (5.95) are equal, or

G(

tb

)2

= π2E6

(bL

)2

(5.96)

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or

Lb

= bt

√π2E6G

= bt

√π2(1 + v)

3(5.97)

as G = E/2(1 + v).

The transition from yield to torsional buckling occurs when the stress in Eq. (5.94)equals the yield stress σ0, or

bt

=√

Gσ0

=√

E2(1 + ν)σ0

(5.98)

and from yield to Euler buckling when the stress in Eq. (5.95) equals σ0, or

Lb

=√

π2E6σ0

(5.99)

Note, other sections such as thin walled angles may also suffer similar behaviour, butthe analysis is more complex as buckling is about the principal axes.

5.2.2 Torsional Buckling Interaction

The critical axial load for torsional buckling Ncr is given in Chapman and Buhagiar(1993) as

Ncr,T = 1i2s

(GIt + π2EIw

l2T

)(5.100)

where lT is the buckling length, and is the polar radius of gyration given by

i2s = i2z + i2y + y20 + z2

0 (5.101)

where iz and iy are the flexural radii of gyration and z0 and y0 are distances from theshear centre to the geometric centroid. For a section whose centroid and shear centreco-incide y0 and z0 are zero. Alternatively the critical stress σcr,T is given by

σcr,T = 1I0

(GIt + π2EIw

l2T

)(5.102)

where I0 is the polar moment of area.

Timoshenko and Gere (1961) give the following interaction equation between strutbuckling and torsion buckling

i2s (N − Ncr,z)(N − Ncr,y)(N − Ncr,T ) − N2z20(N − Ncr,y) − N2y2

0(N − Ncr,z) = 0

(5.103)

where N is the critical value of the axial load and Ncr,z and Ncr,y are the Euler bucklingloads about the zz and xx axes.

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For a section whose centroid and shear centre co-incide, Eq. (5.103) reduces to

(N − Ncr,z)(N − Ncr,y)(N − Ncr,T ) = 0 (5.104)

That is, N is therefore the least of Ncr,x, Ncr,y and Ncr,T .

Chapman and Buhagiar (1993) also indicate that where buckling can occur about bothaxes (i.e. where the buckling lengths and second moments of area are approximatelyequal, or where the critical loads are similar), then the imperfection factor η in thestandard strut buckling interaction equation should be taken as twice its normal valueto allow for torsional buckling.

5.3 PLATE GIRDERS

Plate girders are used either on long spans where a rolled section would need to bespliced and as a result may be inefficient, or to support heavy loads such as on a bridgestructure. It is important to note that although plate girders may be lighter than otherforms of compound beams, fabrication costs are likely to be much higher. Also asCorus now roll UKB’s with a depth of 1016 mm, the use of plate girders in buildingstructures unless spans are extremely high is less likely.

Plate girders are built up from two flange plates and a web plate, generally from thesame grade of steel. Continuous automatic electric arc or submerged gas welding isused to form the fillet welds between the flange and web (Fig. 5.20). Such welding isgenerally performed as a double pass one on either side of the girder. It should benoted that this process may cause very high residual stresses to exist in the flanges andweb. The exact magnitudes will depend also on whether the flange plates were shearedor flame cut to size (Nethercot, 1974a).

The stiffeners are then welded in place often manually. Such stiffeners are neededeither to help combat the effects of web buckling or to provide support to any concen-trated load or reaction. Only straight girders with equal flanges and vertical stiffenersare considered in this text.

Flange plate

Web plateFillet welds

1st pass 2nd pass

Welds from1st pass

Welds on2nd pass

FIGURE 5.20 Plate girder fabrication

Ch05-H5060.tex 12/7/2007 15: 15 page 133


5.3.1 Minimum Web Thickness

With no web stiffeners the web should be sized to avoid the flange undergoing localbuckling due to the web being unable to support the flange. This is known as flangeinduced buckling (cl. 8 EN 1993-1-5).

In pure bending the flanges are subjected to equal and opposite forces. The force perunit length is given by

σvtw = fyfAfc

R(5.105)

where σv is the vertical stress in the web, Afc is the area of the compression flange andR is the radius of curvature. The curvature is dependant upon the variation of strain�εf occurring at the mid-depth of the flanges. The residual strain due to fabricationεf is assumed to have a value of 0,5εy, and �εf is the sum of the yield strain εy and theresidual strain εf , that is, a total of 1,5εy. The radius of curvature R is then given by

R = 0,5hw

�εf(5.106)

where for simplicity hw has been taken as the depth between the centroids of the flangesrather than the clear depth of the web. It should be noted any error will be small.

From Eqs (5.105) and (5.106),

σv = 3Afc

Aw

f 2yf

E(5.107)

The stress σv cannot exceed the elastic critical buckling stress for a simply supportedthin plate which is given by (Bulson, 1970) as

σv = π2E12(1 − ν2)

(twhw

)2

(5.108)

Equating Eqs (5.107) and (5.108) gives with a slight change in notation with Ac

replacing Afc

hw

tw≤ 0,55

Efyf

√Aw

Ac(5.109)

EN 1993-1-5 modifies the co-efficient of 0,55 to allow for situations where higherstrains are required. Thus the critical hw/tw ratio is given by

hw

tw≤ k

Efyf

√Aw

Ac(5.110)

where fyf is the yield strength of the compression flange, Afc is the effective area ofthe compression flange and Aw is the area of the web. The parameter k takes valuesof 0,3 where plastic hinge rotation is utilized, 0,4 if the plastic resistance is utilized

Ch05-H5060.tex 12/7/2007 15: 15 page 134


and 0,55 if the elastic resistance is utilized. Thus for rigid (continuous) design k = 0,3unless the analysis is elastic with no redistribution. For simply supported beams k maybe taken as 0,4.

5.3.2 Bending Resistance

The section classification is determined in the same manner as rolled sections.

5.3.2.1 Compression Flange Restrained (i.e. Lateral TorsionalBuckling cannot Occur)

There are two methods that can be used for girder design:

(1) The flanges carrying the bending moment and the web the shear forceThis is probably best used where the maximum bending moment and maximumshear force are not coincident, as the ability of the flange to contribute towardsshear capacity may be utilized. Thus for a restrained beam under a UDL, thismethod may be advantageous. If the maximum bending moment and maximumshear are coincident in either a simply supported beam under point loading or ina continuous beam at the internal support, then the flange capacity will not beable to be utilized to resist shear, thus probably necessitating a thicker web.With this method, the moment capacity is only dependant on the sectionclassification of the flanges as the web does not carry compression.

(2) The girder carrying the forces as an entityThis method of design is more complex as the beam is likely to be Class 4 and maynot show any resultant economies over the first method, but should be utilizedwhere maximum moment and maximum shear are co-incident.

5.3.2.2 Lateral Buckling May Occur

In this case the second method must be used. The design then follows that for rolledbeams, except that the general method for calculatingχLT should be used with the valueof αLT appropriate for welded sections. Section properties will need to be calculatedfrom first principles.

5.3.3 Basic Dimensioning

One method of dimensioning is to consider a minimum weight solution. It must benoted that a minimum weight solution is not synonymous with a minimum cost solution(Gibbons, 1995).

Method 1:

Assuming the moment to be resisted by the flanges alone, then

MRd = fydbf tfhw (5.111)

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where fyd is the design strength of the flanges, tf and bf the thickness and width of theflange plates and hw the distance between the internal faces of the flanges. Equation(5.111) is slightly conservative for beams of Classes 1 to 3.

The cross-sectional area A is given by

A = 2bf tf + hwt (5.112)

Eliminate bftf between Eqs (5.111) and (5.112) to give

A = 2MRd

hwfyd+ hwt (5.113)

Define the web slenderness ratio hw/t as λ, then Eq. (5.113) becomes

A = 2MRd

λtfyd+ λt2 (5.114)

For an optimum solution, dA/dt = 0, so Eq. (5.114) becomes,

dAdt

= −2MRd

λfydt2 + 2λt = 0 (5.115)

or,

t = 3

√MRd

λ2fyd(5.116)

and

hw = 3

√λMRd

fyd(5.117)

The area of the web, Aw is then given by

Aw = 3

√√√√M2Rd

λf 2yd

(5.118)

Using Eq. (5.112), the flange area, Af is given by

Af = bf tf = MRd

fyd3

√λMRd

fyd

= 3

√√√√M2Rd

λf 2yd

(5.119)

Thus the area of a single flange is equal to that of the web.

Method 2 (Classes 1 and 2):

It is recognized that this is not a likely case but is included for completeness. Themoment is resisted by the complete section, when the moment capacity is given bythat due to the flanges (Eq. (5.111)) and the additional plastic capacity of the web

MRd = fydbf tfhw + fydth2

w4

(5.120)

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where fyd is the design strength of the flanges, tf and bf the thickness and width of theflange plates and hw the distance between the internal faces of the flanges.

The cross-sectional area A is given by Eqs (5.112), thus from Eqs (5.112) and (5.120),bf tf is given by

bf tf = MRd

fydhw− hwt

4(5.121)

Eliminate bf tf between Eqs (5.112) and (5.121) to give

A = 2MRd

hwfyd+ hwt − 2

hwt4

= 2MRd

hwfyd+ hwt

2(5.122)


A = 2MRd

λtfyd+ λt2

2(5.123)


dAdt

= −2MRd

λfydt2 + λt = 0 (5.124)

or,

t = 3

√2MRd

λ2fyd(5.125)

and

hw = 3

√2λMRd

fyd(5.126)


Aw = 3

√√√√4M2Rd

λf 2yd

(5.127)


Af = bf tf = 3

√√√√M2Rd

λf 2yd

[3

√12

− 3

√116

](5.128)

or,

Af

Aw=

3

√M2

Rdλf 2

yd

[3√

12 − 3

√116

]

3

√4M2

Rdλf 2

yd

= 14

(5.129)

Thus the area of the web is equal four times that of a single flange.

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Method 2 (Class 3):

The moment is resisted by the complete section, when the moment capacity is givenby that due to the flanges (Eq. (5.111)) and the additional elastic capacity of the web

MRd = fydbf tfhw + fydth2

w6

(5.130)

where fyd is the design strength of the flanges, tf and bf the thickness and width of theflange plates and hw the distance between the internal faces of the flanges.

The cross-sectional area A is given by Eq. (5.112), thus from Eqs (5.112) and (5.130),bf tf is given by

bf tf = MRd

fydhw− hwt

6(5.131)

Eliminate bf tf between Eqs (5.112) and (5.131) to give

A = 2MRd

hwfyd+ hwt − 2

hwt6

= 2MRd

hwfyd+ 2hwt

3(5.132)


A = 2MRd

λtfyd+ 2λt2

3(5.133)


dAdt

= −2MRd

λfydt2 + 43λt = 0 (5.134)

or,

t = 3

√3MRd

2λ2fyd(5.135)

and

hw = 3

√3λMRd

2fyd(5.136)


Aw = 3

√√√√9M2Rd

4λf 2yd

(5.137)


Af = bf tf = 3

√√√√M2Rd

λf 2yd

[3

√23

− 16

3

√94

](5.138)

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Thus

Af

Aw=

3

√M2

Rdλf 2

yd

[3√

23 − 1

63√

94

]

3

√9M2

Rd4λfyd

= 12

(5.139)

Thus the area of the web is equal twice that of a single flange.

It should be noted that a plate girder may well be Class 4 in which case an effective sec-tion needs calculating, and thus minimum weight optimization is not directly possible,although it can be simulated by increasing the applied moment (see Example 5.4).

5.3.4 Web Design

Experimental work (Basler, 1961; Porter et al., 1975; Rockey et al., 1978; Davies andGriffith, 1999) showed that after web buckling occurred there was still a reserve ofstrength in the web. This additional reserve of strength in the web is due to a tensionfield forming in the central diagonal portion of the web (Fig. 5.21).

The shear capacity in the web is determined using a shear buckling slenderness λw

which is dependant upon the critical shear strength τcr (cl. A.1 EN 1993-1-5)

The critical shear strength τcr is given by

τcr = kτσE (5.140)

where kτ is a shear buckling co-efficient dependant upon the aspect ratio of a webpanel and σE is the elastic critical stress,

From classical plate buckling theory (Bulson, 1970)

σE = π2E12(1 − ν2)

(t

hw

)2

= 19000(

thw

)2

(5.141)

where t is the thickness and hw the depth.

The parameter kτ is given as (Bulson, 1970),

for a/hw ≥ 1,0

kτ = 5,34 + 4,00(

hw

a

)2

(5.142)

for a/hw < 1,0

kτ = 4,00 + 5,34(

hw

a

)2

(5.143)

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a

h w

(b) Panel showing tension field

(a) Buckled shape

τcr

τ cr

Tension panel

Buckled panel

Buckled panel

τ

Stiffener

FIGURE 5.21 Shear failure of a plate girder

The non-dimensionalized web slenderness ratio λw is defined as (cl. 5.3 EN 1993-1-5)

λw =⎡⎢⎣

fyw√3

τcr

⎤⎥⎦

1/2

= 0,76

√fyw

τcr(5.144)

For webs with transverse stiffeners at the supports and either intermediate transverseor longitudinal stiffeners, the normalized web slenderness ratio λw is obtained by

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substituting Eqs (5.140) and (5.141) into (5.144)

λw = hw

37,4εt√

kτ

(5.145)

For webs with transverse stiffeners only at the supports, a/hw is large, hence with littleloss in accuracy kτ = 5,34 from Eq. (5.142). Thus λw is given by

λw = hw

37,4εt√

5,34= hw

86,4εt(5.146)

Clearly the upper limit to the value the shear force that may be carried by the web is theplastic shear capacity. The limiting value of web slenderness beyond which bucklingneed to be considered may be derived as follows. From Eq. (5.142) for an infinitelylong web, kI = 5,34, thus from Eq. (5.140), the limiting value of hw/t is given whenτcr = τyw = fyk/

√3, or

hw

t=√

5,34π2E235 × 12(1 − ν2)

√235fyk

=√

5,34π2 × 210 × 103

12(1 − 0,32) × 235ε = 65,7ε (5.147)

The code uses a lower limit of 72ε/η for unstiffened webs and 31ε√

kτ /η (cl 5.1.(2), EN1993-1-5). The recommended value of η for steel grades of S460 or lower is 1,2. Forsteel grades higher than S460, η = 1,0. With η = 1,2, the lower limit for steel gradesup to and including S460 is 72ε/1,2 = 60ε which is slightly more conservative than thefigure derived in Eq. (5.147).

Basler (1961) and Rockey and Škaloud (1971) recognized that actual web behaviourcould be categorized by three regimes: pure shear, elastic buckling at the extremesand a transition phase between the two limits.

(1) Non-rigid end postIn the case of a non-rigid end post (which cannot generate post-buckling strengthas it is incapable of resisting the additional horizontal forces) EN 1993-1-5 onlydefines two zones for the contribution of the web to shear buckling resistance χw

where a tension field cannot be generated as the anchorage force is unable to besustained,

λw ≤ 0,83/η

χw = η (5.148)

λw > 0,83/η

χw = 0,83λw

(5.149)

(2) Rigid end postWhere there is a rigid end stiffener, the anchorage force from tension field theorycan be sustained. The original theory behind tension field theory was outlined byPorter et al. (1975) and Rockey et al. (1978). EN 1993-1-5 has adopted Höglund’srotating stress–field theory (Davies and Griffith, 1999) which is easier to apply

Ch05-H5060.tex 12/7/2007 15: 15 page 141


than the original tension field theory. Höglund’s rotating stress–field theory alsomobilizes post-buckling behaviour, but only if there is a rigid end post. In this casethe relationships between normalized web slenderness λw and χw are given by

λw ≤ 0, 83/η

χw = η (5.150)

0,83/η ≤ λw < 1,08

χw = 0,83λw

(5.151)

λw ≥ 1,08

χw = 1,370,7 + λw

(5.152)

It will be observed that the difference between the two methods is that the web shearparameter χw is enhanced for λw ≥ 1,08.

The design resistance of a web Vb,Rd whether stiffened or unstiffened (cl 5.2 EN1993-1-5) is given by

Vb,Rd = χvhwt

fyw√3

γM1(5.153)

where shear co-efficient χv is given by

χv = χw + χf ≤ η (5.154)

The parameter χf represents the contribution to shear resistance from the flanges forMEd < Mf,Rd and is given by

χf = bf t2f fyf

√3

cthwfyw

[1 −

(MEd

Mf,Rd

)2]

(5.155)

where bf is the width of the flange taken as not greater than 15εtf on each side of theweb, tf is the web thickness, Mf,Rd is design moment of resistance of the cross-sectiondetermined using the effective flanges only and c is the width of the portion of the webbetween the plastic hinges (see Fig. 5.22) and is given by

c = a(

0,25 + 1,6Mpl,f

Mpl,w

)= a

(0,25 + 1,6bf t2

f fyf

th2wfyw

)(5.156)

Equation (5.155) is derived by determining the shear that may be carried by the portionof both flanges of length c with the yield strength fyf reduced by considering the effectof induced axial forces in the flanges.

The background to the simplified method adopted in the Code is given in Davies andGriffith (1999).

Ch05-H5060.tex 12/7/2007 15: 15 page 142


a

(a) Basic geometry of web panel

(b) Failure mechanism used to derive post buckling strength of web

Web

Position of hinges Flange

d

c

FIGURE 5.22 Overall behaviour of web at failure

The calculations for web shear capacity are iterative as both the shear factors and theapplied moment are dependant upon the stiffener spacing. The calculations for shearcapacity in the examples following were performed on a spreadsheet.

5.3.5 Stiffeners

5.3.5.1 Rigid End Post (cl. 9.3.1 EN 1993-1-5)

This may either be a set of flats welded at the end and above the support with thecentroids a distance e apart, or the end post may comprise a rolled section in whichcase e is the distance between the flange centroids.

The horizontal stress σh for large slenderness ratios can be given as

σh = 0.43λw

fy (5.157)

Substitute the value of λw from Eq. (5.145) into Eq. (5.157) to give an upper boundfor the UDL qh as

qh = σht = 16.1fyt2ε

√kτ

hw(5.158)

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Taking the maximum value of ε as 1,0 and the maximum value of kτ as 9,34, the valueof qh becomes

qh = 49t2fyhw

(5.159)

The distributed load is not uniform over the depth of the girder and the theoreticalvalue of σh is high, the co-efficient of 49 in Eq. (5.159) is replaced by 32.

Assuming the end post is simply supported, the maximum moment Mmax is given by

Mmax = qhh2w

8(5.160)

The section modulus, ignoring any contribution from the web, W is given by Amine,thus assuming the maximum stress is given by fy, then

σmax = fy = Mmax

W=

32t2fyhw

h2w8

Amine(5.161)

Thus the minimum cross-sectional area of each pair Amin is given by

Amin = 4h2wt

e(5.162)

with e > 0,1hw.

The restriction on e would appear to be a detailing requirement.

The end panel may be designed as non-rigid shear panel carrying the whole of theapplied shear. This was originally proposed by Basler (1961).

From Eq. (5.153) with χf = 0, the required value of χw is given by

χw = VEd√

3γM1

fywhwt(5.163)

From Eq. (5.149), the normalized web slenderness λw is given by

λw = 0,83χw

(5.164)

The buckling parameter kτ is given from Eq. (5.145) as

kτ =(

hw

37,4tελw

)2

(5.165)

As a/hw < 1, kτ is given by Eq. (5.143), or the required panel width a is given as

a = hw

√5,35

kτ − 4,00(5.166)

Note, this will give an upper bound to the value of a.

Ch05-H5060.tex 12/7/2007 15: 15 page 144


5.3.5.2 Transverse Stiffeners (cl 9.2.1 EN 1993-1-5)

These should be checked as a simply supported beam with an initial sinusoidal imper-fection w0 given by Eq. (5.169) together with any eccentricities. The transverse stiffenershould carry the deviation forces from the adjacent panels assuming the adjacent trans-verse stiffeners are rigid and straight. A second order analysis should be used to deter-mine that the maximum stress does not exceed fyd nor any additional deflection b/300.This will be more critical for single sided stiffeners. As double sided plate stiffenershave been used in the ensuing examples, only the stress criterion has been checked.

In the absence of transverse loads or axial forces in the stiffeners, then the strengthand deflection criteria are satisfied if they have a second moment of area given by

Ist = σM

E

(bπ

)4 (1 + w0

300b

u)

(5.167)

where σM is given by

σM = σcr,c

σcr,p

NEd

b

(1a1

+ 1a2

)(5.168)

where a1 and a2 are the panel lengths either side of the stiffener under consideration,NEd is the larger compressive force in the adjacent panels, b is the height of the stiffener.

The initial imperfection w0 is given as

w0 = 1300

LEAST(a1, a2, b) (5.169)

The parameter u is given by

u = π2Eemax300bfyγM1

≥ 1,0 (5.170)

The distance emax is taken from the extreme fibre of the stiffener to the centroid ofthe stiffener.

The critical stress for plate between vertical stiffeners σcr,c is given by

σcr,c = π2Et2

12(1 − ν2)a2 (5.171)

The critical stress σcr,p is given by kσ,pσE. The value of σE is given by Eq. (5.141). Thevalue for plates with longitudinal stiffeners in Annex A of EN 1993-1-5. For unstiffenedplates σcr,p = σcr,c

To avoid lateral torsional buckling of the stiffener,

IT

Ip≥ 5,3

fyE

(5.172)

where IT and Ip are St. Venant torsional constant for the stiffener alone and Ip is thepolar second moment of area about the edge fixed to the plate. Eq. (5.172) can be

Ch05-H5060.tex 12/7/2007 15: 15 page 145


derived as follows. The critical buckling stress σcr for open section stiffeners (withnegligible warping stiffness) is given by Eq. (5.102) with Iw = 0,

σcr = GIT

Ip(5.173)

From cl 9.2.1(8) of EN 1993-1-5, the critical stress σcr is limited by

σcr ≥ θfy (5.174)

For plate stiffeners, θ is taken as 2,0, thus Eq. (5.173) becomes

IT

Ip≥ 5,2

fyE

(5.175)

The code replaces the co-efficient 5,2 by 5,3.

5.3.5.3 Intermediate Transverse Stiffeners (cl. 9.3.3 EN 1993-1-5)

The force Ns,Rd to be resisted by a stiffener is given by

Ns,Rd = VEd − χwhwt

fyw√3

γM1(5.176)

Note, χw is calculated for the web panel between adjacent stiffeners assuming thestiffener under consideration is removed. In the case of variable shear, then the checkis performed at a distance 0,5hw from the edge of the pane with the larger shear force.

To determine the buckling resistance of the stiffener a portion of the web may takeninto account (Rockey et al., 1981). A section of the web in length equal to 15εt eitherside of the stiffener may be considered (cl 9.1, EN 1933-1-3) (Fig. 5.23)

For a symmetric stiffener, the effective area Ae is given by

Aequiv = Ast + 30εt2 (5.177)

and the effective second moment of area Iequiv by

Iequiv = Ist + 112

30εt4 (5.178)

where Ast is the area of the stiffener and Ist is the second moment of area of thestiffener. For end stiffeners the co-efficient of 30 in Eqs (5.177) and (5.178) should bereplaced by 15.

The effective length of the stiffener may be taken as 0,75hw and buckling curve ‘c’ usedto determine the strength reduction factor (cl 9.4 EN 1993-1-5).

In order to provide adequate restraint against buckling it was found that the stiffenersneed to possess a minimum second moment of area (Rockey et al., 1981).

Ch05-H5060.tex 12/7/2007 15: 15 page 146


15 εt 15 εt

t

Section AA

A A

Elevation

FIGURE 5.23 Stiffener geometry

The minimum second moment of area Is is given by

for a/hw <√

2

Ist ≥ 1,5h3

wt3

a2 (5.179)

for a/hw ≥ √2

Ist ≥ 0,75hwt3 (5.180)

It can be demonstrated that for compression buckling a change from 1 to 2 half sinewaves occurs at a/hw = √

2, and that thereafter the buckling co-efficient is sensiblyindependent of the aspect ratio of the panel. Thus Eq. (5.180) is determined fromEq. (5.179) by substituting a = hw

√2 (cl 9.3.3 (3) EN 1993-1-5).

5.3.5.4 Plate Splices (cl 9.2.3 EN 1993-1-5)

The splice whether in the web or flanges, should ideally occur at a transverse stiffener.If not then the stiffener should be at a distance no greater than b0/2 along the thinnerplate where b0 is the depth of the web (or the least spacing of longitudinal stiffeners).

5.3.5.5 Longitudinal Welds (Web to Flange) (cl 9.3.5 EN 1993-1-5)

The weld between the web and flange(s) should be designed for a shear flow of VEd/hw,provided

VEd < χwhwt

fyw√3

γM1(5.181)

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If the condition in Eq. (5.181) is not satisfied, the welds should be designed under ashear flow of ηt(fyw/

√3)/γM1.

EXAMPLE 5.3 Design of a laterally restrained plate girder

Design a plate girder in Grade S355 steel to carry a characteristic variable load of150 kN/m over a span of 20 m. The compression flange is fully restrained againstlateral torsional buckling.

For a plate girder span/depth ratios are generally around 8 or 10 to 1. The higherthis ratio is the lower the flange size but at the probable expense of a thicker web toovercome buckling.

To calculate the imposed bending moment, assume the total weight of the beam is100 kN.

Total loading on the beam = 1,5 × 150 + 1,35 × (100/20) = 232 kN/m

MSd = 232 × 202/8 = 11,6 MNm

The critical slenderness ratio for the web can be controlled by flange-induced bucklingwith k = 0,4 as plastic rotation is not utilized. If the flanges resist the bending moment,Aw = Af (from Eq. (5.119)), thus the critical hw/t ratio is given by Eq. (5.110)

hw

t≤ k

Efyf

√Aw

Ac= 0,4

210 × 103

355

√1 = 237

From Eq. (5.117) calculate hw

hw = 3

√λMRd

fyd=√√√√237 × 11,6 × 109

3551,0

= 1978 mm

The thickness t is given as

t = hw

λ= 1978

237= 8,34

Use t = 9 mm

Maximum ratio flange outstand to flange thickness beyond the weld for a Class 1section is 9ε where ε = (235/355)1/2 (=0,814).

So maximum flange outstand is 9 × tf × (235/355)1/2 = 7,33tf

So flange area is 14,66t2f (ignoring the effect of weld width).

Mpl,Rd ≈ Affy

γM0hw

Ch05-H5060.tex 12/7/2007 15: 15 page 148


or

Af = MEd

hwfy

γM0

= 11,6 × 109

1978 3551,0

= 16, 520 mm2

or

14,66t2f = 16,520

or

Inset tf = 33,6 mm

Use 35 mm thick plate with a width of 500 mm.

Overall depth, h:

h = 2 × 35 + 1978 = 2048 mm. Use h = 2000 mm.

Check actual hw/t ratio:

Actual web slenderness:

hw

t= 1930

9= 214

Web is Class 4. However as the webs do not carry any compression, then the section maybe treated dependant upon the classification of the flanges (cl 5.5.2 (12) EN 1993-1-1).

Maximum hw/t ratio:

hw

t= k

Efyf

√Aw

Af= 0,4

210 × 103

355

√1930 × 9500 × 35

= 236

The actual value is below the allowable, and is therefore satisfactory.

Plastic moment of resistance of the flanges, Mpl,Rd:

Mpl,Rd = Affy

γM0(h − tf ) = 35 × 500

3551,0

(2000 − 35) = 12,2 MNm

Vpl,Rd = 232 × 10 = 2320 kN (at the support)

hw = h − 2tf = 2000 − 2 × 35 = 1930 mm

The web will be designed both ways, non-rigid and rigid end post:

The determined flange and web contributions take the maximum shear and maximummoment in a web panel, even though they are at opposite ends of the panel. This willbe conservative.

Determination of χf :

To determine χf , the flange width is limited to 15εtf on either side of the web:

15εtf = 15 × 35(

235355

)1/2

= 427 mm

Ch05-H5060.tex 12/7/2007 15: 15 page 149


Actual flange width = 0,5(500 − 9) = 245 mm. This is less, therefore use actual width,so Mf,Rd = 12,2 MNm (from above).

Both the web and flanges have a yield strength of 355 MPa.

Non-rigid end post:

First panel from the support:

Intermediate stiffener is 1,2 m from the support.

Determine c from Eq. (5.156)

c = a

(0,25 + 1,6bf t2

f fyf

th2wfyw

)= 1200

(0,25 + 1,6 × 500 × 352 × 355

9 × 19302 × 355

)= 335 mm

MEd = 2320 × 1,2 − 2321,22

2= 2617 kNm

Determine the flange contribution factor χf from Eq. (5.155)

χf = bf t2f fyf

√3

cthwfyw

[1 −

(MEd

Mf,Rd

)2]

= 500 × 352 × 355 × √3

335 × 9 × 1930 × 355

[1 −

(261712200

)2]

= 0,174

hw

a= 1930

1200= 1,608

ahw

= 0,622

For a/hw < 1,0, kτ is given by Eq. (5.143), or

kτ = 4 + 5, 34(

hw

a

)2

= 4 + 5,34 × 1,6082 = 17,81

Determine the normalized web slenderness ratio λw from Eq. (5.145)

λw = hw

37,4εt√

kτ

= 1930

37,4√

235355 × 9

√17,81

= 1,670

As λw > 1,08, χw is given by Eq. (5.149) as

χw = 0,83λw

= 0,831,670

= 0,497

From Eq. (5.154), the total shear co-efficient χv is given by

χv = χf + χw = 0,174 + 0,497 = 0,671

Ch05-H5060.tex 12/7/2007 15: 15 page 150


TABLE 5.5 Shear capacity calculations with non-rigid end post (Example 5.3).

Distance from Support (m) 1,2 2,55 4,15 6,25 10Panel width (m) 1,2 1,35 1,6 2,1 3,75Moment MEd (kNm) 2617 5162 7630 9969 11600Shear V Ed (kN) 2320 2042 1728 1357 870c (mm) 335 377 447 586 1047χf 0,174 0,133 0,083 0,034 0,006a/hw 0,622 0,699 0,829 1,088 1,943kτ 17,81 14,91 11,77 8,719 6,400λw 1,670 1,825 2,054 2,387 2,786χw 0,497 0,455 0,404 0,378 0,298χv 0,671 0,588 0,487 0,382 0,304V Rd (kN) 2389 2093 1735 1361 1081

The shear capacity Vb,Rd is determined from Eq. (5.153)

Vb,Rd = χvhwt

fyw√3

γM1= 0,671 × 1930 × 9

1000

355√3

1,0= 2389 kN

The calculations for subsequent panels are summarized in Table 5.5 (together with thefirst panel)

Intermediate stiffeners:

Check strength:

First stiffener:

The axial force NEd is given by Eq. (5.176) as

NRd = VEd − χwhwt

fyw√3

γM1

As the load is a UDL, VEd is determined at 0,5 hw from the stiffener in the panel withthe higher shear:

VEd = 2320 − 232(1,2 − 0,5 × 1,930) = 2265 kN

The web contribution parameter χw is calculated assuming the stiffener is removed,thus a = 1200 + 1350 = 2550 mm

hw

a= 1930

2550= 0,757

ahw

= 1,321

For a/hw > 1,0, kτ is given by Eq. (5.142), or

kτ = 5,34 + 4(

hw

a

)2

= 5,34 + 4 × 0,7572 = 7,63

Ch05-H5060.tex 12/7/2007 15: 15 page 151



λw = hw

37,4εt√

kτ

= 1930

37,4 × 9√

235355

√7,63

= 2,55


χw = 0,83λw

= 0,832,55

= 0,325


fyw√3

γM1= 2265 − 0,325 × 1930 × 9 × 10−3

355√3

1,0= 1108 kN

Second stiffener:

Axial force NEd is given by Eq. (5.176) as


fyw√3

γM1

As the load is a UDL, VEd is determined at 0,5hw from the stiffener in the panel withthe higher shear:

VEd = 2320 − 232(2,55 − 0,5 × 1,930) = 1952 kN


hw

a= 1930

2950= 0,654

ahw

= 1,528


kτ = 5,34 + 4(

hw

a

)2

= 5,34 + 4 × 0,6542 = 7,05


λw = hw

37,4εt√

kτ

= 1930

37,4 × 9√

235355

√7,05

= 2,65


χw = 0,83λw

= 0,832,65

= 0,313


fyw√3

γM1= 1952 − 0,313 × 1930 × 9 × 10−3

355√3

1,0= 838 kN

Ch05-H5060.tex 12/7/2007 15: 15 page 152


Third stiffener:

Axial force NEd is given by Eq. (5.176) as


fyw√3

γM1


VEd = 2320 − 232(4,15 − 0,5 × 1,930) = 1581 kN


hw

a= 1930

3700= 0,522

ahw

= 1,917


kτ = 5,34 + 4(

hw

a

)2

= 5,34 + 4 × 0,5222 = 6,43


λw = hw

37,4εt√

kτ

= 1930

37,4 × 9√

235355

√6,43

= 2,78


χw = 0,83λw

= 0,832,78

= 0,299


fyw√3

γM1= 1581 − 0,299 × 1930 × 9 × 10−3

355√3

1,0= 517 kN

Fourth stiffener:

Axial force NEd is given by Eq. (5.176)


fyw√3

γM1


VEd = 2320 − 232(6,25 − 0,5 × 1,930) = 1094 kN

Ch05-H5060.tex 12/7/2007 15: 15 page 153



hw

a= 1930

5850= 0,330

ahw

= 3,03


kτ = 5,34 + 4(

hw

a

)2

= 5,34 + 4 × 0,332 = 5,78


λw = hw

37,4εt√

kτ

= 1930

37,4 × 9√

235355

√5,78

= 2,93


χw = 0,83λw

= 0,832,93

= 0,283


fyw√3

γM1= 1094 − 0,283 × 1930 × 9 × 10−3

355√3

1,0= 86 kN

This value is small, thus there is no need to check the centre stiffener.

The minimum stiffness requirement for all but the two panels either side of the centralstiffener is given by the case a <

√2hw, so design on the least value of a:

Is = 1,5h3

wt3

a2 = 1,519303 × 93

12002 = 5,46 × 106 mm4

Use 9 mm thick plate, then the total breadth of the stiffener b is given by

b = 3

√12 × 5,46 × 106

9= 194 mm

Use b = 200 mm. The axial force that can be carried by the stiffener NRd is given as

NRd = 2 × 200 × 93551,0

× 10−3 = 1278 kN.

Buckling check:

Effective length = 0,75 × 1930 = 1448 mm

From Eq. (5.177), the effective area Aequiv is given as

Aequiv = Ast + 30εt2 = 200 × 9 + 30 × 9

√235355

= 2020 mm2

Ch05-H5060.tex 12/7/2007 15: 15 page 154


From Eq. (5.178), the effective area Iequiv is given as

Iequiv = Ist + 30εt4 = 112

2003 × 9 + 112

30 × 94

√235355

= 6,013 × 106 mm4

Ncr = π2EIequiv

L2 = π2 × 210 × 6,013 × 106

14482 = 5944 kN

λ =√

AequivfyNcr

=√

2020 × 355 × 10−3

5944= 0,121

As λ ≤ 0,2 strut buckling need not be checked (cl 6.3.1.2 (4) (EN 1993-1-1)).

Thus the minimum stiffener size will be adequate for all the intermediate stiffeners.

End stiffener:

Try a 500 wide by 15 thick plate:

The section must be checked for buckling, but a proportion of the web may be takeninto account.

Length of web = 15εt = 15 × 9√

(235/355) = 110 mm

As,eff = As + Aweb = 500 × 15 + 110 × 9 = 8490 mm2

Is,eff = Is + Iweb = 15 × 5003

12+ 110 × 93

12= 0,156 × 109 mm4

i =√

Is,eff

As,eff=√

0,156 × 109

8490= 135,6 mm

Use an effective length of 0,75hw = 0,75 × 1930 = 1448 mm

Ncr = π2EIl2

= 210 × 0,156 × 109π2

14482 = 154210 kN

λ =√

AfyNcr

=√

8490 × 355154210 × 103 = 0,14

From cl 6.3.1.2 (4) (EN 1993-1-1), there is no reduction for strut buckling as λ < 0,2.

So,

NRd = Afy

γM0= 8490

3551.0

× 10−3 = 3014 kN

This exceeds the reaction of 2320 kN.

Check cl 9.2.1 (7)

Determine Ip:

The second moment of area about the web centre line, Iy:

Iy = bh3

12= 15 × 5003

12= 0,156 × 109 mm4

Ch05-H5060.tex 12/7/2007 15: 15 page 155


The second moment of area normal to the web centre line about one edge, Iy:

Iy = bh3

3= 500 × 153

3= 0,563 × 106 mm4

Ip = Ix + Iy = 0,156 × 109 + 0,563 × 106 = 0,157 × 109 mm4

IT = bh3

3= 0,563 × 106 mm4

IT

Ip= 0,563 × 106

0,157 × 109 = 3,59 × 10−3

Limiting value:

5,3fyE

= 5,3355

210 × 103 = 8,96 × 10−3

The actual value is less than the limiting value thus the stiffener size must be increased.

A 25 mm thick end plate will satisfy the limiting stiffness criterion (IT/IP = 0,01) (andwill clearly satisfy the strength and buckling criteria).

Flange to web welds:

From Table 5.5 VEd > hwt( fyw/√

3)/γM1 as the flange contribution χf has beenmobilized. Thus welds should be designed for a shear flow of

ηt

fyw√3

γM1= 1,2 × 9

355√3

1,0= 2214 N/mm

The final layout of the girder whose self-weight is 91,1 kN is given in Fig 5.24(a).

Rigid end post:


Intermediate stiffener is 1,45 m from the support.


c = a

(0,25 + 1,6bf t2

f fyf

th2wfyw

)= 1450

(0,25 + 1,6 × 500 × 352 × 355

9 × 19302 × 355

)

= 405 mm

MEd = 2320 × 1,45 − 2321,452

2= 3120 kNm


χf = bf t2f fyf

√3

cthwfyw

[1 −

(MEd

Mf,Rd

)2]

= 500 × 352 × 355 × √3

405 × 9 × 1930 × 355

[1 −

(312012200

)2]

= 0,141

Ch05-H5060.tex 12/7/2007 15: 15 page 156


(a) Non-rigid end post

(b) Rigid end post

2,0

1,22,55

4,15

6,25

Intermediate stiffeners

2 � 200 � 9

End stiffeners500 � 25

35 � 500 flanges9 mm web

cL

2,0

1,45

3,25

6,0

0,3

Intermediate2 � 200 � 9

End post2 � 500 � 15

cL

FIGURE 5.24 Final layout for the beam in Example 5.3

hw

a= 1930

1450= 1,331

ahw

= 0,751


kτ = 4 + 5,34(

hw

a

)2

= 4 + 5,34 × 1,3312 = 13,46


λw = hw

37,4εt√

kτ

= 1930

37,4√

235355 × 9

√13,46

= 1,920


χw = 1,370,7 + λw

= 1,370,7 + 1,92

= 0,523


χv = χf + χw = 0,141 + 0,523 = 0,664

Ch05-H5060.tex 12/7/2007 15: 15 page 157



Vb,Rd = χvhwt

fyw√3

γM1= 0,664 × 1930 × 9

1000

355√3

1,0= 2363 kN

The resultant spacing of stiffeners and a summary of the remaining calculations isgiven in Table 5.6 The final result for the calculation of stiffener forces are also givenin Table 5.6.

The minimum stiffness requirement for all but the two panels either side of the centralstiffener is given by the case a <


Is = 1,5h3

wt3

a2 = 1,519303 × 93

14502 = 3,74 × 106 mm4


b = 3

√12 × 3,74 × 106

9= 171 mm

Use b = 200 mm. The axial force that can be carried by the stiffener NRd is given as

NRd = 2 × 200 × 93551,0

× 10−3 = 1278 kN.

This exceeds the values of NEd in the last line of Table 5.6.

Rigid end post:

Use Eqs (5.163)–(5.166) to determine the maximum panel size in accordance with cl9.3.1(4) (EN 1993-1-5)

Determine χw from Eq. (5.163)

χw = VEd√

3γM1

fywhwt= 2320 × 103

√3 × 1,0

355 × 1930 × 9= 0,652

TABLE 5.6 Shear capacity calculations with rigid end post (Example 5.3).

Distance from support (m) 1,45 3,25 6,0 10Panel width (m) 1,45 1,80 2,75 4,00Moment MEd (kNm) 3120 6315 9744 11 600Shear V Ed (kN) 2320 1984 1566 928c (mm) 405 503 768 1117χf 0,141 0,089 0,029 0,005a/hw 0,751 0,933 1,425 2,073kτ 13,46 10,14 7,310 6,271λw 1,920 2,213 2,607 2,814χw 0,523 0,470 0,414 0,390χv 0,664 0,559 0,443 0,395V Rd (kN) 2363 1991 1578 1407Stiffener force N Rd (kN) 1118 758 154 −826

Ch05-H5060.tex 12/7/2007 15: 15 page 158


The required normalized web slenderness λw from Eq. (5.164) is given by

λw = 0,83χw

= 0,830,652

= 1,273

The buckling parameter kτ is given by Eq. (5.165) as

kτ =(

hw

37,4tελw

)2

=⎛⎜⎝ 1930

37,4 × 9 ×√

235355 × 1,273

⎞⎟⎠

2

= 30,65


a = hw

√5,35

kτ − 4,00= 1930

√5,35

30,65 − 4= 865 mm

Centre to centre distance of the pair of double stiffeners should be greater than 0,1 hw

(=193 mm). Use a centre to centre distance of 300 mm (The shear resistance will begreater than 2320 kN).

Minimum cross-sectional area of each plate is

4hwt2

e= 4 × 1930 × 92

300= 2084 mm2

As the beam is 500 mm wide use an end plate 500 wide, thus the required thickness is2084/500 (=4 mm). As for the non-rigid end post, a stiffener 500 mm by 15 mm sufficesfor resisting the reaction, use the same here.

The layout of the beam whose self-weight is 93 kN is given in Fig 5.24 (b). The self-weight per unit length of the beam is 4,56 kN/m for the non-rigid end post and 4,51for the rigid end post (i.e. almost identical). It would be possible to further optimizethe weight of the girder by decreasing the web thickness towards the centre and theflange width and/or thickness towards the ends of the beam. Note, however, that sodoing may also increase the stiffener requirements.

EXAMPLE 5.4 Design of a plate girder with lateral torsional buckling.

Design a plate girder in Grade S355 steel to carry a characteristic variable load of1500 kN at the centre of a span of 20 m. Lateral torsional restraints exist at the supportsand the load.

To calculate the imposed bending moment, assume the total weight of the beam is150 kN.

MEd = 1,35 × 150 × 208

+ 1,5 × 1500 × 204

= 11760 kNm

Ch05-H5060.tex 12/7/2007 15: 15 page 159


To size the beam allowing for the fact that its classification will be Class 4 (needing thedetermination of an effective section) and lateral torsion buckling will occur, increasethe moment by 50% and use the optimization equations for Class 3.

MEd = 1,5 × 1500 × 204

+ 1,35 × 200 × 208

= 11925 kNm

Approximate design:

As the moment resistance is calculated on an elastic resistance utilization, the factork in Eq. (5.110) may be taken as 0,55, with the flange area equal to half the web areafor an optimal design under elastic stress distribution (Eq. (5.139)), so

λ = hw

t= k

Efyf

√Aw

Af= 0,55

210 × 103

355

√2 = 460

To determine the optimal solution increase MEd by 50% to 1,5 × 11925 (=17888 kNm).

From Eq. (5.135), t is given by

t = 3

√3MEd

2λ2fyd= 3

√3 × 17888 × 106

2 × 4602 × 355= 7,1 mm

Use 8 mm plate.

Determine hw from Eq. (5.136)

hw =√

3λMEd

2fyd=√

3 × 460 × 17888 × 106

2 × 355= 3264 mm

As the plate thickness has been rounded up from 7,1 to 8 mm, then the web heightmay be rounded down to 3240 mm.

The area of the flange Af is given by Eq. (5.139)

Af = Aw

2= 8 × 3240

2= 12960 mm2

Use the limit of the flange outstand as a Class 1 section (i.e. 7,33tf ) so the area of theflange is 14,66t2

f . Thus

14,66t2f = 12960, or tf = 29,7 mm

Use a flange plate 30 mm thick.

bf ≈ 14,66tf = 14,66 × 30 = 440 mm

Use a flange plate 500 mm wide (Af = 15000 mm2).

The reason for keeping the flange classification as low as possible is in order to mobilizeas much of the flange capacity as possible to resist shear.

Ch05-H5060.tex 12/7/2007 15: 15 page 160


Actual web slenderness

hw

t= 3240

8= 405

Allowable web slenderness:

hw

t= k

Efyf

√Aw

Af= 0,55

210 × 103

355

√3240 × 8500 × 30

= 427

Thus flange-induced buckling will not occur. From the actual web slenderness the web(and therefore the complete section) is Class 4.

Determination of effective section properties (cl 4.4 EN 1993-1-5):

The process is iterative as the amount of web not considered is a function of thestresses at the top and bottom of the web. Note, the calculation takes compressivestresses positive rather than the normal stress analysis convention of compressivestresses negative.

First Iteration:

Determine the stresses at the top and bottom of the web with no loss of section:

Gross Area, A

A = 2bf tf + hwt = 2 × 500 × 30 + 3240 × 8 = 55920 mm2

Gross Iy:

Iy = 112

[b(hw + 2tf )3 − (b − t)h3w] = 1

12[500 × 33003 − 492 × 32403]

= 102, 88 × 109 mm4

Stress at the top of the web, σ1:

σ1 = 11 925 × 106 × 1620102,88 × 109 = 188 MPa


σ2 = −11 925 × 106 × 1620102,88 × 109 = −188 MPa

Determine the stress ratio, ψ:

ψ = σ2

σ1= −188

188= −1,0

From Table 4.1 of EN 1993-1-5, the buckling factor kσ = 23,9 for ψ = −1,0.

The normalized slenderness ratio λp is given by

λp = b

28,4tε√

kσ

= 3240

28,4 × 8√

235355

√23,9

= 3,585

where b is the depth of the web.

Ch05-H5060.tex 12/7/2007 15: 15 page 161


The reduction factor ρ for an internal compression member is given by

ρ = λp − 0,055(3 + ψ)

(λp)2= 3,585 − 0,055(3 + (−1))

3,5852 = 0,271

The effective depth beff is given by

beff = ρhw

1 − ψ= 0,271 × 3240

1 − (−1)= 439 mm

The depth of web left at the top be1:

be1 = 0,4beff = 0,4 × 439 = 176 mm

The depth of web left at the bottom (above the centroidal axis), be2:

be1 = 0,6beff = 0,6 × 439 = 263 mm

The ineffective portion of web has a length lw = 1620 − 176 − 263 = 1181 mm

The net loss of area of the web, Aw is given by

Aw = 1181 × 8 = 9448 mm2

Effective area of section, Aeff :

Aeff = A − Aw = 59920 − 9448 = 50472 mm2

Position of effective centroid, zeff :

zeff =A h

2 − Aw

(zeff,prev + be2 + lw

2

)Aeff

=59920 3300

2 − 9448(

33002 + 263 + 1181

2

)50472

= 1490 mm

Note, zeff,prev is the neutral axis position at the previous iteration. For the first iteration,zeff,prev = h/2.

Effective second moment of area, Iy,eff :

Iy,eff = Iy + A(

h2

− zeff

)2

− tl3w12

− Aw


2− zeff

)2

= 102,88 × 109 + 59920(

33002

− 1490)2

− 8 × 11813

12

− 9448(

33002

+ 263 + 11812

− 1490)2

= 102,88 × 109 + 1,53 × 109 − 1,10 × 109 − 9,70 × 109 = 93,61 × 109 mm4

Ch05-H5060.tex 12/7/2007 15: 15 page 162



σ1 = 11925 × 106 × (3270 − 1490)93,61 × 109 = 227 MPa

Stress at the bottom of the web, σ2:

σ2 = −11925 × 106 × (1490 − 30)93,61 × 109 = −186 MPa

Determine the stress ratio, ψ:

ψ = σ2

σ1= −186

227= −0,819

From Table 4.1 of EN 1993-1-5, the buckling factor kσ for ψ = −0,794 is given by

kσ = 7,81 − 6,29ψ + 9,78ψ2 = 7,81 − 6,29(−0,819) + 9,78(−0,819)2 = 19,5

The normalized slenderness ratio λp is given by

λp = b

28,4tε√

kσ

= 3240

28,4 × 8√

235355

√19,5

= 3,97

where b is the depth of the web.

The reduction factor ρ for an internal compression member is given by

ρ = λp − 0,055(3 + ψ)

(λp)2= 3,97 − 0,055(3 + (−0,819))

3,972 = 0,244

The effective depth beff is given by

beff = ρhw

1 − ψ= 0,244 × 3240

1 − (−0,819)= 435 mm

The depth of web left at the top be1:

be1 = 0,4beff = 0,4 × 435 = 174 mm

The depth of web left at the bottom (above the centroidal axis), be2:

be1 = 0,6beff = 0,6 × 435 = 261 mm

The ineffective portion of web has a length lw = (3240 − 1490) − 174 − 261 = 1315 mm

The net loss of area of the web, Aw is given by

Aw = 1315 × 8 = 10520 mm2

Effective area of section, Aeff :

Aeff = A − Aw = 59920 − 10520 = 49400 mm2

Ch05-H5060.tex 12/7/2007 15: 15 page 163


Position of effective centroid, zeff :

zeff =A h

2 − Aw


2

)Aeff

=59920 3300

2 − 10520(

1490 + 261 + 13152

)49400

= 1488 mm

Effective second moment of area, Iy,eff :

Iy,eff = Iy + A(

h2

− zeff

)2

− tl3w12

− Aw


2− zeff

)2

= 102,88 × 109 + 59920(

33002

− 1488)2

− 8 × 13153

12

− 10520(

1490 + 261 + 13152

− 1488)2

= 102,88 × 109 + 1,57 × 109 − 1,52 × 109 − 8,91 × 109 = 94,02 × 109 mm4


σ1 = 11925 × 106 × (3270 − 1488)94,02 × 109 = 226 MPa

Stress at the bottom of the web, σ2:

σ2 = −11925 × 106 × (1488 − 30)94,02 × 109 = −185 MPa

As these stresses are virtually identical to those on the first iteration, there is no needto continue.

Note, that as the changes in be1 and be2 were small the iteration could have stoppedwithout further calculations of σ1 and σ2 (ψ = −0,819)

The lesser elastic section modulus Weff,y is given as

Weff,y = Ieff

zeff − tf= 94,02 × 109

1488 − 30= 64,49 × 106 mm3

MRd = Weff,yfy

γM0= 64,49 × 106 355

1,0× 10−6 = 22900 kNm

Lateral torsional buckling check

Moment gradient factor, C1 from Eq. (5.40), with ψ = 0:

1C1

= 0,4ψ + 0,6 = 0,6

Ch05-H5060.tex 12/7/2007 15: 15 page 164


The gross section is used to calculate the section properties required for Mcr.

Iz = 112

[2tfb3f + hwt3] = 1

12[2 × 0,030 × 0,53 + 3,24 × 0,0083]

= 0,625 × 10−3 m4

Iw = Izh2

s4

= 0,625 × 10−3 3,2702

4= 1.67 × 10−3 m6

Note: hs is the distance between the centroids of the flanges.

The torsional second moment of area may be calculated on the thin plate assumption.

It = 13

[2bf t3f + hwt3] = 1

3[2 × 0,5 × 0,0303 + 3,24 × 0,0083] = 9,55 × 10−6 m4

Use Eq. (5.5) to determine Mcr

Mcr = π2EIz

L2

[Iw

Iz+ L2GIt

π2EIz

]1/2

L = 10 m.

π2EIz

L2 = π2 × 210 × 106 × 0,625 × 10−3

102 = 12954 kN

GIt = 81 × 106 × 9,55 × 10−6 = 774 kNm2

GIt

π2EIzL2

= 77412954

= 0,060 m2

Iw

Iz= 1,67 × 10−3

0,625 × 10−3 = 2,672 m2

Mcr = 12954[2,672 + 0,060]1/2 = 21410 kNm

From Eq. (5.12), λLT is given by

λLT =√

fyWeff,y

C1Mcr=√

0,6355 × 64,49 × 106 × 10−6

21410= 0,801

h/b > 2, so αLT = 0,76

�LT = 0,5[1 + αLT(λLT − 0,2) + (λLT)2]

= 0,5[1 + 0,76(0,801 − 0,2) + 0,8012] = 1,049

χLT = 1�LT + (�2

LT − (λLT)2)1/2= 1

1,049 + (1,0492 − 0,8012)1/2 = 0,579

Mb,Rd = χLTWeff,yfy

γM1= 0,579 × 64,49 × 106 355

1,0× 10−6 = 13260 kNm

This exceeds the applied moment of 11925 kN.

The beam is therefore satisfactory for flexure.

Ch05-H5060.tex 12/7/2007 15: 15 page 165


Deflection check:

Igross = 102,88 × 109 mm4

For a worst case scenario assume that all the variable load contributes to the deflection,

δ = 148

WL3

EI= 1

481500 × 203

210 × 106 × 102,88 × 10−3 = 0,012 m

This is equivalent to span/1667, which is satisfactory.

Web design:

As the web carries compression due to flexure, then the following interaction equationmust be satisfied (cl 7.1(1), EN 1993-1-5),

η1 +(

1 − Mf,Rd

Mpl,Rd

)(2η3 − 1)2 ≤ 1,0 (5.182)

where Mf,Rd is the plastic moment resistance of the flanges and Mpl,Rd is the plasticmoment of resistance of the section (both calculations are irrespective of classificationof the section). The shear contribution factor η3 is defined as (cl. 5.5 (1), EN 1993-1-5)

η3 = VEd

χvhwt[(fyw/√

3)/γM1]≤ 1, 0 (5.183)

as there is no axial force the equation for η1 from cl 4.6 (1) (EN 1993-1-5) reduces to

η1 = MedfyWeff

γM0

≤ 1,0 (5.184)

If η3 < 0,5 there is no reduction in moment capacity (cl 7.1 (1), EN 1993-1-5)

Mpl,Rd = Wplfy

γM0= 500 × 33002 − 492 × 32402

43551,0

× 10−6 = 24870 kNm

Mf,Rd = Afhsfy

γM0= 500 × 30 × (3300 − 30)

3551,0

× 10−6 = 17410 kNm

(1 − Mf,Rd

Mpl,Rd

)= 1 − 17410

24870= 0,3

Flexible end post:

Three intermediate stiffeners are required at 2,8, 5,6 and 7,8 m from the support,together with a load bearing stiffener at the centre.

Stiffener at 2,8 m:

MEd = 1260 × 2,8 − 6,75 × 2,82 = 3475 kNm

VEd = 1260 kN (at support)

Ch05-H5060.tex 12/7/2007 15: 15 page 166



c = a

(0,25 + 1,6bf t2

f fyf

th2wfyw

)= 2800

(0,25 + 1,6 × 500 × 302 × 355

8 × 32402 × 355

)= 724 mm


χf = bf t2f fyf

√3

cthwfyw

[1 −

(MEd

Mf,Rd

)2]

= 500 × 302 × 355 × √3

724 × 8 × 3240 × 355

[1 −

(347517410

)2]

= 0,040

hw

a= 3240

2800= 1,157

ahw

= 0,864


kτ = 4 + 5,34(

hw

a

)2

= 4 + 5,34 × 1,1572 = 11,15


λw = hw

37.4εt√

kτ

= 3240

37.4√

235355 × 8

√11.15

= 3.985


χw = 0,83λw

= 0,833,985

= 0,208


χv = χf + χw = 0,040 + 0,208 = 0,248


Vb,Rd = χvhwt

fyw√3

γM1= 0,248 × 3240 × 8

1000

355√3

1,0= 1318 kN

η3 = VEd

χvhwt[(fyw/√

3)/γM1]= 1260

1318= 0,956

As η3 > 0,5 the interaction equation between moment and shear must be considered(Eq. (5.152))

η1 = MedfyWeff

γM0

= 347522900

= 0,152

Ch05-H5060.tex 12/7/2007 15: 15 page 167


TABLE 5.7 Shear capacity calculations with non-rigid end post (Example 5.4).

Distance from support (m) 2,8 5,6 7,8 10Panel width (m) 2,8 2,8 2,2 2,2Moment MEd (kNm) 3475 6844 9417 11 925Shear V Ed (kN) 1260 1241 1222 1207c (mm) 724 724 568 568χf 0,040 0,035 0,038 0,028a/hw 0,864 0,864 0,679 0,679kτ 11,15 11,15 15,58 15,58λw 3,990 3,990 3,371 3,371χw 0,208 0,208 0,246 0,246χv 0,248 0,243 0,284 0,274V Rd (kN) 1318 1293 1506 1457η1 0,152 0,209 0,411 0,521η3 0,956 0,960 0,811 0,829Interaction equation 0,401 0,553 0,527 0,650Stiffener force (kN) 1079 210 −134 −932

η1 +(

1 − Mf,Rd

Mpl,Rd

)(2η3 − 1)2 = 0,152 + 0,3(2 × 0,956 − 1)2 = 0,402 ≤ 1,0

The calculations for the remaining stiffeners are carried out in Table 5.7, together withthe calculation of the forces on the stiffeners.

Intermediate stiffeners:

For all the panels, a ≤ √2hw (=3,24

√2 = 4,58 m), thus based on the lesser value of a

(=2,2 m) the stiffener requirement, Ist is given by

Ist = 1,5h3

wt3

a2 = 1,532403 × 83

22002 = 5,4 × 106 mm4

Use 8 mm plate, so

b = 3

√12 × 5,4 × 106

8= 200 mm

Use double intermediate stiffeners of 200 × 8 mm plates either side of the web.

Load capacity of stiffeners NRd (with no buckling):

NRd = Astfy

γM0= 2 × 200 × 8

3551,0

× 10−3 = 1136 kN

This exceeds the maximum stiffener force of 1079 kN.

As in earlier examples strut buckling is not critical, it will not be checked in thisexample.

End post:

VEd = 1260 kN

Use a 500 wide by 10 mm thick plate.

Ch05-H5060.tex 12/7/2007 15: 15 page 168




(235/355) = 98 mm

From Eq. (5.177) Aequiv is given by

Aequiv = Ast + 15εt2 = 500 × 10 + 98 × 8 = 5784 mm2

and Iequiv by

Iequiv = Ist + 112

15εt4 = 10 × 5003

12+ 98 × 83

12= 0,104 × 109 mm4

i =√

Iequiv

Aequiv=√

0,104 × 109

5784= 134,1 mm

Use an effective length of 0,75hw = 0,75 × 3240 = 2430 mm

Ncr = π2EIl2

= 210 × 0,104 × 109π2

24302 = 36500 kN

λ =√

AfyNcr

=√

5784 × 35536500 × 103 = 0,237

Use buckling curve ‘c’ (α = 0,49)

� = 0,5[1 + α(λ − 0,2) + (λ)2] = 0,5[1 + 0,49(0,237 − 0,2) + 0,2372] = 0,537

χ = 1

� +√

�2 − (λ)2= 1

0,537 +√0,5372 − 0,2372= 0,981

NRd = χAfy

γM0= 0,981 × 5784

3551,0

× 10−3 = 2014 kN

This exceeds the reaction of 1260 kN.

Check cl 9.2.1 (7)

Determine Ip:

The second moment of area about the web centre line, Iy:

Iy = bh3

12= 10 × 5003

12= 0,104 × 109 mm4

The second moment of area normal to the web centre line about one edge, Iy:

Iy = bh3

3= 500 × 103

3= 0,042 × 106 mm4

Ip = Ix + Iy = 0,104 × 109 + 0,042 × 106 = 0,104 × 109 mm4

IT = bh3

3= 0,042 × 106mm4

IT

Ip= 0,042 × 106

0,104 × 109 = 0,4 × 10−3

Ch05-H5060.tex 12/7/2007 15: 15 page 169


Limiting value:

5,3fyE

= 5,3355

210 × 103 = 8,96 × 10−3

The actual value is less than the limiting value thus the stiffener size must be increased.

A 25 mm thick end plate will satisfy the limiting stiffness criterion (IT/IP = 0,01) (andwill clearly satisfy the strength and buckling criteria).

Central stiffener:

NEd = 1,5 × 1500 = 2250 kN

Use two 240 wide by 15 mm thick plates.



(235/355) = 195 mm

Aequiv = Ast + 30εt2 = 2 × 240 × 15 + 195 × 8 = 8760 mm2

Isequiv = Ist + 30εt4 = 15 × (240 + 8 + 240)3

12+ 195 × 83

12= 0,145 × 109 mm4

i =√

Iequiv

Aequiv=√

0,145 × 109

8760= 128,7 mm

Use an effective length of 0,75 hw = 0,75 × 3240 = 2430 mm

Ncr = π2EIl2

= 210 × 0,145 × 109π2

24302 = 50895 kN

λ =√

AfyNcr

=√

8760 × 35550895 × 103 = 0,247

Use buckling curve ‘c’ (α = 0,49):

� = 0,5[1 + α(λ − 0,2) + (λ)2] = 0,5[1 + 0,49(0,247 − 0,2) + 0,2472] = 0,542

χ = 1

� +√

�2 − (λ)2= 1

0,542 +√0,5422 − 0,2472= 0,976

NRd = χAfy

γM0= 0,976 × 8760

3551,0

× 10−3 = 3035 kN

This exceeds the applied load of 2250 kN.

Ch05-H5060.tex 12/7/2007 15: 15 page 170


(a) Non-rigid end post

(b) Rigid end post

cL

Intermediate 2 � 200 � 8

End 500 � 25

500 � 30 flanges8 mm web

cL

3,3

2,8

5,6

7,8

Central 2 � 240 � 15

5,00,4

Intermediate2 � 150 � 8

End post500 � 25

Central 2 � 240 � 15

FIGURE 5.25 Final layout for the girder of Example 5.4

Flange to web welds:

From Table 5.7, VEd > hwt[(fyw/√

3)/γM1] as the flange contribution χf has beenmobilized. Thus welds should be designed for a shear flow of

ηt

fyw√3

γM1= 1.2 × 8

355√3

1.0= 1968 N/mm

The final layout is given in Fig 5.25 (a). The self-weight of the beam is 101 kN (or5,05 kN/m).

(b) Rigid end post


Intermediate stiffener is 5 m from the support.

MEd = 1260 × 5 − 6,75 × 52 = 6131 kNm

VEd = 1260 kN (at the support)

Ch05-H5060.tex 12/7/2007 15: 15 page 171



c = a

(0,25 + 1,6bf t2

f fyf

th2wfyw

)= 5000

(0,25 + 1,6 × 500 × 302 × 355

8 × 32402 × 355

)= 1293 mm


χf = bf t2f fyf

√3

cthwfyw

[1 −

(MEd

Mf,Rd

)2]

= 500 × 302 × 355 × √3

1293 × 8 × 3240 × 355

[1 −

(613117410

)2]

= 0,020

hw

a= 3240

5000= 0,648

ahw

= 1,543


kτ = 5,34 + 4(

hw

a

)2

= 5,34 + 4 × 0,6482 = 7,02


λw = hw

37,4εt√

kτ

= 3240

37,4√

235355 × 8

√7,02

= 5,02


χw = 1,370,7 + λw

= 1,370,7 + 5,02

= 0,240


χv = χf + χw = 0,020 + 0,240 = 0,260


Vb,Rd = χvhwt

fyw√3

γM1= 0,260 × 3240 × 8

1000

355√3

1,0= 1380 kN

η3 = VEd

χvhwt[(fyw/√

3)/γM1]= 1260

1380= 0,913

Ch05-H5060.tex 12/7/2007 15: 15 page 172


As η3 > 0,5 an interaction equation between moment and shear must be considered,

η1 +(

1 − Mf,Rd

Mpl,Rd

)(2η3 − 1)2 ≤ 1,0

η1 = MedfyWeff

γM0

= 613122900

= 0,268

η1 +(

1 − Mf,Rd

Mpl,Rd

)(2η3 − 1)2 = 0,268 + 0,3(2 × 0,913 − 1)2 = 0,473 ≤ 1,0

The resultant spacing of stiffeners and a summary of the remaining calculations isgiven in Table 5.8. The final result for the calculation of stiffener forces are also givenin Table 5.8.

Note, it is possible to eliminate the intermediate stiffener by increasing the web thick-ness to 9 mm. This may be more economic when considering the overall materials andfabrication costs.

The minimum stiffness requirement for both panels either side of the central stiffeneris given by the case a >


Ist = 0,75hwt2 = 0,75 × 3240 × 92 = 0,197 × 106 mm4


b = 3

√12 × 0,197 × 106

8= 67 mm

Minimum area to carry a stiffener force of 757 kN is given as

A = 757 × 103

355= 2132 mm2

TABLE 5.8 Shear capacity calculations with rigid end post (Example 5,4).

Distance from support (m) 5 10Panel width (m) 5 5Moment MEd (kNm) 6131 11 925Shear V Ed (kN) 1260 1226c (mm) 1293 1293χf 0,020 0,012a/hw 1,543 1,543kτ 7,02 7,02λw 5,02 5,02χw 0,239 0,239χv 0,259 0,261V Rd (kN) 1380 1337η1 0,268 0,521η3 0,913 0,917Interaction equation 0,473 0,729Stiffener force N Rd (kN) 757 −419

Ch05-H5060.tex 12/7/2007 15: 15 page 173


This would require a total width of 2132/8 = 267 mm. Use two plates 150 mm by 8 mmas intermediate stiffeners.

Design of rigid end post.

Use Eqs (5.163) to (5.166) to determine the maximum panel size in accordance withcl 9.3.1(4) (EN 1993-1-5)

Determine χw from Eq. (5.163)

χw = VEd√

3γM1

fywhwt= 1260 × 103

√3 × 1.0

355 × 3240 × 8= 0,237

The required normalized web slenderness λw from Eq. (5.164) is given by

λw = 0,83χw

= 0,830,237

= 3,502

The buckling parameter kτ is given by Eq. (5.165) as

kτ =(

hw

37,4tελw

)2

=⎛⎜⎝ 3240

37,4 × 8 ×√

235355 × 3,502

⎞⎟⎠

2

= 11,44


a = hw

√5,35

kτ − 4,00= 3240

√5,35

11,44 − 4= 2750 mm

Centre to centre distance of the pair of double stiffeners should be greater than 0.1 hw

(=324 mm). Use a centre to centre distance of 400 mm (the shear resistance will begreater than 1260 kN).

Minimum cross-sectional area of each plate is

4hwt2

e= 4 × 3240 × 82

400= 2074 mm2

As the beam is 500 mm wide use an end plate 500 wide, thus the required thickness is2074/500 (=4 mm).

As for the non-rigid end post, a stiffener 500 mm by 25 mm is needed.

Stiffener under the point load:

The calculations are the same as those for the non-rigid end post.

The final layout is given in Fig 5.25 (b). The self-weight is 105 kN (5,06 kN/m)

Ch05-H5060.tex 12/7/2007 15: 15 page 174


REFERENCES

Anderson, J.M. and Trahair, N.S. (1972). Stability of mono-symmetric beams and cantilevers, Journalof the Structural Division, Proceedings of the American Society of Civil Engineers, 98, 269–286.

Basler, K. (1961). Strength of plate girders in shear, Journal of the Structural Division, Proceedings ofthe American Society of Civil Engineers, 87, 151–180.

Bradford, M.A. (1989). Buckling of beams supported on seats, Structural Engineer, 69(23), 411–414.Bulson, P.S. (1970). The stability of flat plates. Chatto and Windus.Chapman, J.C. and Buhagiar, D. (1993). Application of Young’s buckling equation to design against

torsional buckling. Proceedings of the Institution of Civil Engineers, Structures and Buildings, 99,359–369.

Davies, A.W. and Griffiths, D.S.C. (1999). Shear strength of steel plate girders, Proceedings of theInstitution of Civil Engineers, 134, 147–157.

Gibbons, C. (1995). Economic steelwork design, Structural Engineer, 73(15), 250–253.Kaim, P. (2006). Buckling of members with rectangular hollow sections, In Tubular structures XI (eds

Packer and Willibald). Taylor and Francis Group, 443–449.Kirby, P.A. and Nethercot, D.A. (1979). Design for structural stability. Granada.Mutton, B.R. and Trahair, N.S. (1973). Stiffness requirements for lateral bracing, Journal of the

Structural Division, ASCE, 99, 2167.Nethercot, D.A. (1974a). Residual stresses and their influence upon the lateral buckling of rolled

steel beams, Structural Engineer, 52(3), 89–96.Nethercot, D.A. (1974b). Buckling of welded beams and girders, IABSE, 34, 163–182.Nethercot, D.A.and Lawson, R.M. (1992). Lateral stability of steel beams and columns – common cases

of restraint, Publication 093. Steel Construction Institute.Nethercot, D.A. and Rockey, K.C. (1971). A unified approach to the elastic critical buckling of beams,

Structural Engineer, 49(7), 321–330.Pillinger, A.H. (1988). Structural steelwork: a flexible approach to the design of joints in simple

construction, Structural Engineer, 66(19), 316–321.Porter, D.M., Rockey, K.C. and Evans, H.R. (1975). The collapse of plate girders loaded in shear,

Structural Engineer, 53(8), 313–325.Rees, D.W.A. (1990). Mechanics of solids and structures. McGraw Hill.Rockey, K.C. and Škaloud, M. (1971). The ultimate load behaviour of plate girders loaded in shear,

IABSE Colloquium – design of plate girders for ultimate strength. London, 1–19.Rockey, K.C., Evans, H.R. and Porter, D.M. (1978). A design method for predicting the collapse

behaviour of plate girders, Proceedings of The Institution of Civil Engineers, 65, 85–112.Rockey, K.C., Valtinat, G. and Tang, K.H. (1981). The design of transverse stiffeners on webs loaded

in shear – an ultimate load approach, Proceedings of The Institution of Civil Engineers, 71(2),1069–1099.

Rondal, J., Würker, K-G., Dutta, D., Wardenier, J. and Yeomans, N. (1992). Structural stability ofhollow sections. Verlag TÜV Rheinland.

Timoshenko, S.T. and Gere, J.M. (1961). Theory of elastic stability. McGraw Hill.Trahair, N.S. (1983). Lateral buckling of overhanging beams, In Instability and plastic collapse of steel

structures (ed L.J. Morris). Granada, 503–518.Trahair, N.S. and Bradford, M.A. (1988). The behaviour and design of steel structures (2nd Edition).

Chapman and Hall.BS 5950-1: Structural use of steelwork in building – Part 1: Code of practice for design – Rolled and

welded sections. BSI.EN 1993-1-1. Eurocode 3: Design of steel structures, Part 1-1: General rules and rules for buildings.

CEN/BSI.EN 1993-1-5. Eurocode 3: Design of steel structures, Part 1-5: Plated structural elements. CEN/BSI.

Ch06-H5060.tex 11/7/2007 14: 10 page 175

C h a p t e r 6 / Axially LoadedMembers

6.1 AXIALLY LOADED TENSION MEMBERS

A member subject to axial tension extends and tends to remain straight or, if there is asmall initial curvature, to straighten out as the axial load is increased. Tension members(ties) occur in trusses, bracing and hangers for floor beams. A flat can be used as atie, but this is generally impractical because it buckles if it goes into compression. Tiesections are therefore angles and tees for small loads and ‘I’ sections for larger loads.In situations where the load is not applied axially then the member is designed to resistan axial force plus a bending moment.

A tension member extends when subject to an axial load and is deemed to have failedwhen the yield or ultimate stress is reached. The failure load is independent of thelength of the member which is in contrast to an axially loaded compression memberwhich fails by buckling.

A member which is purely in tension does not buckle locally or overall and is thereforenot affected by the classification of sections. The characteristic stress is not reduced,for design purposes, except by the material factor.

6.1.1 Angles as Tension Members (cl 4.13, EN 1993-1-8 (2005))

Generally angles are connected by one leg at the end of the member and this intro-duces an eccentric load. Where an angle is connected to one side of a gusset plate(as in a truss) bending moments are introduced in addition to the direct axial force.For a tension member these moments produce lateral deflections which reduce theeccentricity of the load near the middle of the member. Thus under increasing loadthe bending stresses become concentrated more towards the ends of the member.For angles connected by one leg the principal sectional axes are inclined to the planecontaining the bending moment. Secondary deflections therefore occur normal to theplane of bending and, because of the restraints provided by the gusset plates, twistingalso takes place.

Generally eccentrically loaded members are designed to resist an axial load and bend-ing moment. However angle and tee experiments (Nelson (1953); Regan and Salter

Ch06-H5060.tex 11/7/2007 14: 10 page 176

176 • Chapter 6 / Axially Loaded Members

(1984)) demonstrated that the above effects could be compensated for in design byreducing the cross-sectional area of the member. If there are holes then these alsoreduce the area of the cross-section.

Angles may be treated as axially loaded members provided that the net area is reducedto the effective area (cl 4.13, EN 1993-1-8 (2005)). For an equal angle, or an unequalangle connected by the larger leg the effective area is the gross area. For an unequalangle connected by the smaller leg the effective area is twice that of the smaller leg.

6.1.2 Design Value of a Tension Member (cl 6.2.3,EN 1993-1-1 (2005))

The design value of the tensile force NEd at each cross-section should satisfy (Eq. (6.5),EN 1993-1-1 (2005))

NEd

Nt,Rd≤ 1 (6.1)

where Nt,Rd is the design tension resistance taken as the smaller of:

(a) The design plastic resistance of the gross cross-section (Eq. (6.6), EN 1993-1-1(2005))

Npl,Rd = AfyγM0

(6.2a)

(b) The design ultimate resistance of the net cross-section at holes for fasteners(Eq. (6.7), EN 1993-1-1 (2005))

Nu,Rd = 0,9Anet fuγM2

(6.2b)

(c) The design ultimate resistance of the net cross-section at holes for fasteners whichare preloaded or non-preloaded (Eq. (6.8), EN 1993-1-1 (2005) and cl 3.4.2.1(1),EN 1993-1-8 (2005))

Nnet,Rd = Anet fyγM0

(6.2c)

EXAMPLE 6.1 An angle in tension connected by one leg. A 100 × 65 × 6 mm singleangle tie is connected through the smaller leg by two 20 mm diameter bolts in line witha pitch of 2,5 d0. Determine the design ultimate resistance of the angle assuming S275steel and material factors of γM2 = 1,25 and γM0 = 1.

Net area of angle connected by the smaller leg and allowing for holes

Anet = (b − do)t + bt = (65 − 22) × 6 + 65 × 6 = 648 mm2

Compare this value with the gross area = 1120 mm2 (Section Tables).

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Design ultimate tensile resistance of the net cross-section for a two bolt angleconnection (Eq. (6.7), EN 1993-1-1 (2005))

Nu,Rd = 0,9Anet fuγM2

= 0,9 × 648 × 4301,25 × 1E3

= 200,6 kN.

or

Npl,Rd = AfyγM0

= (2 × 65 × 6) × 2751,0 × 1E3

= 214,5 > 200,6 kN.

A check must be made for the strength of the bolts as shown in Chapter 7.

6.2 COMBINED BENDING AND AXIAL FORCE – EXCLUDING BUCKLING(cl 6.2.9, EN 1993-1-1 (2005))

Combining an axial force (tension or compression) and bending moment inducesstresses which vary across a ductile steel section. At a certain point the stresses maycombine to produce a yield stress but this does not produce collapse of the memberbecause collapse only occurs if the entire section is at yield stress (i.e. the section isplastic). If the axial force is compressive it is assumed that buckling does not occur.

6.2.1 Rectangular Sections (cl 6.2.9.1(3), EN 1993-1-1 (2005))

A rectangular section subject to an axial force and bending moment produces a stressdiagram as shown in Fig. 6.1. The neutral axis is displaced from the equal area positionand the stress diagram can be represented in two parts, one for the axial load and onefor the reduced plastic modulus.

Plastic moment of resistance about y–y axis

My = fybh1(h − h1) (6.3)

Axial load

N = fybh(h − 2h1) (6.4)

h

h 1

b fy

fy fy

fy fy

y y

FIGURE 6.1 Effect of axial force on the plastic moment of resistance of arectangular section

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A

fyb f

t f

tw

fy

2cyy yyh

N

AN

(a) Neutral axis (NA) in web (b) Neutral axis in flange

FIGURE 6.2 Effect of axial force on the plastic moment of resistance of an ‘I’ section

Combining Eqs (6.3) and (6.4) and eliminating h1

My/(bh2fy/4) + [N/(bhfy)]2 = 1 (6.5a)

This relationship is plotted on Fig. 6.3. The term (bh2fy/4) is the plastic moment ofresistance for a rectangular section. This form of the equation is given in Eq. (6.32),EN 1993-1-1 (2005) and is applicable to rectangular solid sections without holes.

MN,Rd = Mpl,Rd

[1 −

(NEd

Npl,Rd

)2]

(6.5b)

6.2.2 ‘I’ Sections

The relationship between bending moment and axial force for an ‘I’ section is morecomplicated. An ‘I’ section subject to an axial load and bending moment produces astress diagram as shown in Fig. 6.2. The neutral axis is displaced from the equal areaposition. This stress diagram can be represented in two parts, one for the axial loadand one for the reduced plastic modulus.

A convenient ratio to determine the reduced plastic section modulus is

n = σ

fy

where

σ is the mean axial stress

fy is the specified minimum yield strength of steel.

If the axial force is small the neutral axis is in the web. Alternatively if the axial forceis large the neutral axis is in the flange as shown in Fig. 6.2. For bending about the y–yaxis, the neutral axis moves from the web into the flange when n > nc where

nc = tw

(h − 2tf

A

)

A is the total area of the section.

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Subtracting the plastic modulus of the hatched area from the whole section.

Wyr = Wy − W(shaded area)

W(shaded area) = tw(2c)2

4

From equilibrium

σA = (2c)twfy

and since by definition n = σ/fy

When n ≤ nc

Wyr = Wy − an2

When n > nc

Wyr = b(1 − n)(c + n)

where

a = A2

4tw

b = A2

4bf

c = 2bfhA

−1

For bending about the y–y axis the change point for n is

nc = twhA

and the values of

a = A2

4h

b = A2

8tf

c = 4tfbf

A−1

Values of a, b, c and nc for ‘I’ sections are given in Section Tables.

The plastic section modulus determined in this way is applicable in tension and forshort columns. For longer columns instability effects due to deflection of the columnreduce this value. The relationship between N/Np and M/Mp for a typical ‘I’ section isplotted in Fig. 6.3.

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M/Mp

N/N

p

1.0

1.0

0FIGURE 6.3 Relationship between N/Np andM/Mp at collapse

The above theory for ‘I’ and ‘H’ sections is simplified in the European Code by thefollowing recommendations (cl 6.2.9.1(4), EN 1993-1-1 (2005)).

(a) For Class 1 and Class 2 cross-sections the plastic moment of resistance is notreduced if axial loads are limited.

Bending about the y–y axis (Eqs (6.33) and (6.44), EN 1993-1-1 (2005))

NEd ≤ 0,25Npl,Rd

NEd ≤ 0,5hwtwfyγM0

Bending about the z–z axis

NEd ≤ hwtwfyγM0

(b) For Class 3 and Class 4 cross-sections the maximum longitudinal stress is limitedto fy/γM0. In the case of Class 4 sections the effective area of the section is used.

6.3 BUCKLING OF AXIALLY LOADED COMPRESSION MEMBERS

Compression members are present in many structures, for example, trusses, bracingand columns. They are generally greater in cross-sectional area than tension membersbecause they may fail in buckling (Fig. 6.4). If buckling is likely to occur then sectionsmust be capable of resisting bending moments.

6.3.1 Compression Members (cl 6.3.1, EN 1993-1-1 (2005))

An efficient cross-section for a strut is a hot finished tube because residual stresses area minimum and the buckling resistance is the same for all axes of bending. However theuse of a tube is not always practical because of the difficulties of making connections.

Ch06-H5060.tex 11/7/2007 14: 10 page 181


a

xy

NE

NE

L/2

L/2

L/2

L/2

(a) Euler strut

x

N

N

y y0

O Δ0

(b) Practical strut

FIGURE 6.4 Bucklingbehaviour based on Eulerstrut and practical strut

Connections to Universal Column ‘I’ sections are simpler but the section is not asefficient because of the weaker z–z axis of bending. Where loads are relatively smallangles and ‘T’ sections are used as struts (e.g. in roof trusses).

The strength of a compression member is not reduced significantly by welding or holeswith fasteners but they must be arranged sensibly.

Short steel compression members fail by squashing at the yield stress, while long, ormore accurately slender members, fail by buckling. Buckling may occur at an axialstress which is less than the yield stress and is related to slenderness ratio, lack ofstraightness and non-axial loads. As buckling progresses the load becomes progres-sively more eccentric to the longitudinal axis of the member and a bending moment isintroduced as shown in Fig. 6.4(a).

6.3.2 Buckling Theory

The Euler theory was the first attempt to produce a rational explanation of bucklingbehaviour of a strut. It was based on the differential equation, related to Fig. 6.4(a),which shows the final deflected form of a pin-ended strut

EId2ydx2 = M = NE(a − y) (6.6)

The solution of Eq. (6.6) shows that an axially loaded pin-ended strut becomeselastically unstable and buckles at the Euler critical stress

fE = π2Eλ2

where

E Young’s elastic modulus

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L length of a pin-ended strut

i radius of gyration

λ slenderness ratio = L/i

The Euler buckling theory makes no allowance for:

(a) homogeneity of column material,(b) isotropy of column material,(c) variation of E value and elastic–plastic behaviour of column material,(d) loading not axial,(e) residual stresses,(f) lack of straightness of a column,(g) cross-section of a column not rectangular,(h) local buckling,(i) alternative end conditions to a column.

These factors are present in practice and reduce the Euler buckling load.

The Euler differential equation can be modified in a number of ways (Bleich (1952))to take account of these factors. One method produces the Perry–Robertson formulawhich is used in the European Code and is related to the Euler buckling stress.

Small unavoidable eccentricities of loading and lack of initial straightness can be sim-ulated mathematically by assuming an initial curvature which produces a small centraldeflection �0 (Fig. 6.4(b)).

When a load N is applied the deflection at x is increased by y and the differentialequation of bending similar to Eq. (6.6) is

EId2ydx2 = M = −N[ y + y0] (6.7)

Adopting a sinusoidal function for the initial curvature

y0 = �0 cos (πx/L) and putting μ2 = N/EI then

d2ydx2 + μ2

[y + �0 cos

(πxL

)]= 0

The solution to this equation is

y = A sin μx + B cos μx + μ2�0 cos (πx/L)π2/L2 − μ2

when x = ±L/2, y = 0 and hence A = B = 0 and

y = μ2�0 cos (πx/L)π2/L2 − μ2

If the Euler buckling load NE = π2EI/L2, then

y = N�0 cos (πx/L)NE − N

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If N/A = f and NE/A = fE then the deflection

y = ffE − f

�0 cos(πx

L

)

and the total deflection at any point

y0 + y =[

fE(fE − f

) + 1

]�0 cos

(πxL

)

The maximum deflection at x = 0

ymax =(

fEfE − f

)�0

and the maximum bending moment is

Mmax =(

fEfE − f

)�0N

If dexf is the distance of the extreme fibre from the neutral axis then the maximumcompressive stress

fexf =(

�0dexfNI

)(f

fE − f

)+ f

If (�0dexfN/I) = (�0dexfAf/I) = (�0dexf/i2)f = ηf where i is the radius of gyration ofthe column section then

fexf = f

[η fE(

fE − f) + 1

]

Assuming that the critical buckling load is reached when yielding commences in theextreme fibres of the strut, that is, when fexf = fy and f = fPR then rearranging thePerry–Robertson buckling stress

fPR = 0,5[ fy + (η + 1) fE] − {0,52[ fy + (η + 1) fE]2 − fy fE}1/2 (6.8)

where the Euler critical buckling stress fE = π2E/λ2.

Equation (6.8) is known as the Perry–Robertson formula and its adoption is explainedby Dwight (1975). The value of the function η has varied over the years and the valueoriginally obtained experimentally by Robertson in 1925 related to the slendernessratio (λ) for circular sections was

η = 0,003λ (6.9)

The value suggested later by Godfrey (1962) to give more economical designs, basedon experimental work by Duthiel in France, was

η = 0,3(

λ

100

)2

(6.10)

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Equation (6.8) has been rearranged in the European Code (Eq. (6.49), EN 1993-1-1(2005)) to express the buckling stress ( fPR) in terms of the stress ratio ( fE/fy = λ) anda reduction factor (χ) related to column imperfections. If

ζ = 0,5[ fy + (η + 1) fE] then

fPR = ς − (ς2 − fy fE)1/2 = fy fE

[ς + (ς2 − fy fE)1/2 ]

= fy[ζ

fE+{(

ζ

fE

)2 −( fy

fE

)}1/2] = χ fy (6.11)

In the European Code ζ/fE = �, fy/fE = λand the reduction factor (χ) is then expressedin terms of λ

� = ζ

fE= [fy/fE + (η + 1)]

2

η = 0,001a (λ − λ0) but not less than zero, where a varies from 2 to 8 depending on theshape of the section and the limiting slenderness ratio

λ0 = 0,2

√(π2E

fy

)> 0

Combining these equations (cl 6.3.1.2, EN 1993-1-1 (2005))

� = 0,5[1 + α(λ − 0,2) + λ2]

where α = 0,001a (π2E/fy)1/2 is an imperfection factor.

For uniform members the buckling stress reduction factor (Eq. (6.49), EN 1993-1-1(2005))

χ = 1/[� + (�2 − λ2)1/2] ≤ 1 (6.12)

The buckling curves are related to the shape of the sections, axis of buckling, andthickness of material as shown in Tables 6.1 and 6.2, and Fig. 6.4, EN 1993-1-1 (2005).The curves are based on experimental results for real columns as described in theECCS (1976) report, and expressed theoretically by Beer and Schultz (1970).

As the lowest value of the central deflection (�) is related to the initial curvature,bending moments are generated as soon as the axial load is applied and the bucklingprocess starts immediately. Therefore there is no condition of elastic instability asdefined by Euler (1759) and the average compressive stress may never reach the Eulercritical stress for a strut of finite length. Nevertheless the failure of a strut is suddenwhen compared with the ductile failure of a tension member.

For a fuller development of the buckling theory see Trahair and Bradford (1988).

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6.3.3 Local Buckling (cl 5.5.2(2) and Table 5.2, EN 1993-1-1(2005))

Slender elements of a section (e.g. flanges), which are primarily in compression, maybuckle locally before overall buckling of the member occurs. This is likely to occurwith Class 4 sections for high values of the width (c) to thickness (t) ratios of elementsof a cross-section. To allow for local buckling the European Code reduces the cross-sectional area to an effective area and consequently the member supports less load.

6.3.4 Column Cross-section

Studies (Bleich, 1952) over the year have shown that the strength of a column isinfluenced by the cross-section. Initial theoretical investigations showed that materialconcentrated at the centre of gravity was more effective in resisting buckling.

Later research identified other factors (Table 6.2, EN 1993-1-1 (2005)) which affectbuckling, for example, hot finished or cold formed, yield strength, buckling axis, weld-ing and shape of cross-section. For design purposes these factors are related to anappropriate buckling curve (Fig. 6.4, EN 1993-1-1 (2005)).

6.3.5 Buckling Length of a Column

The buckling theory, developed previously, is based on the assumption that the endsof the column are pinned, that is, frictionless joints which prevent linear movementbut which can rotate freely about any axis of the section. Pin-ended columns are rarein practice but the pin-ended condition is a useful theoretical concept which can beshown to relate to other end conditions.

Alternative end support conditions can be simulated by replacing the actual lengthof the column by an effective length. Consider the theoretical end conditions shownin Fig. 6.5 which vary from complete end fixity to complete freedom at the end ofa cantilever. The theoretical effective length (l) is expressed as a proportion of theactual length (L) and is the distance between real or theoretical pins (i.e. points ofcontraflexure). The effective length is important in design calculations because theEuler buckling load is inversely proportional to the square of the effective length.

In practice the full rigidity of a fixed built-in end is never achieved and some rotationoccurs due to the flexibility of the connection or the support. Only a small rotationis necessary to transform a built-in end to a pinned end and thus reduce the bucklingresistance of a column. In comparison small translational movements at supports arenot so critical and may be limited by supporting members.

Practical end conditions therefore allow for some rotation and translation at the endsof a real column. For the column in Fig. 6.6 the theoretical effective length is 0,7 Lbut in practice the built-in end can rotate and the real effective length is 0,85 L, which

Ch06-H5060.tex 11/7/2007 14: 10 page 186


L

l = 0

,5L

l = 0

,7L

l = L

l = 2

L

FIGURE 6.5 Theoretical effective lengths

0,7

L

0,85

L

L

Rotation at supportFIGURE 6.6 Rotation of the end of a strut

reduces the axial load at failure. Because perfect rigidity and completely free ends donot occur in practice the range of common practical values is between 0,7 L and 1,5 L.

Methods exist (Williams and Sharp, 1990; Wood, 1974) for determining the effectivelengths for columns in rigid frames which are based on the relative total column stiff-ness at a joint to the total stiffness of all members at the joint. For situations wheresway of the column does not occur, for example, where cross bracing is present insimple construction, the effective length is less than the real length (0,5 L to 1,0 L).Where sway occurs, for example, an unpropped portal frame, the effective length isgreater than the real length (1,0 L to 2,0 L).

For single angles used as struts eccentricity of axial load may generally be ignored. Asa rough guide the effective length is approximately 0,85 L for two fasteners (or weld)

Ch06-H5060.tex 11/7/2007 14: 10 page 187


and 1,0 L for one fastener. Buckling failure about the minor axis must be consideredfor which a single equal angle, is at 45◦ to the major axis of bending. An alternativemethod of determining a modified slenderness ratio for angles is given in cl BB 1.2, EN1993-1-1 (2005). For double angle struts the effective lengths are similar to those fora single angle but failure about the 45◦ axis may not be possible because of restraints.

6.3.6 Maximum Slenderness Ratios

At high values of the slenderness ratio struts become so flexible that deflections undertheir own weight are sufficient to introduce stresses in excess of the Perry–Robertsonbuckling formula. The following empirical limits have been used in the past.

The slenderness ratio (λ) should not generally exceed the following:

(a) for members resisting loads other than wind loads 180(b) for members resisting self-weight and wind loads only 250(c) for any member normally acting as a tie but subject to reversal of

stress resulting from the action of wind 350

Members whose slenderness exceeds 180 should be checked for self-weight deflection.If this exceeds (length/1000) the effect of the bending should be taken into account indesign.

6.3.7 Intermediate Restraints

A member that provides an intermediate restraint and prevents buckling of a strut,reduces the effective length and increases the strength of the strut. The restraint neednot be rigid and may be elastic provided that its stiffness exceeds a certain value (Trahairand Bradford, 1988). Often restraining members associated with built-up members arerequired to resist not less than 1% of the axial force in the restrained member.

6.3.8 Combined Bending, Shear and Axial Force (cl 6.2.10,EN 1993-1-1 (2005))

Th effect of a shear force on the Euler buckling load for solid sections is small (Bleich,1952). However as explained in Chapter 4 the plastic moment of resistance is reducedwhen the design shear force is greater than 50% of the plastic shear resistance. Alsofor large values of shear shear buckling may reduce the resistance of the section (cl 5,EN 1993-1-5 (2005)).

6.3.9 Design Buckling Resistance (cl 6.3.1, EN 1993-1-1 (2005))

Equation (6.8) expresses failure as a buckling stress but the European Code expressesfailure as a design load where local buckling and the buckling stress factor (χ) reducethe buckling load.

Ch06-H5060.tex 11/7/2007 14: 10 page 188


The design buckling resistance of a compression member for Classes 1, 2 and 3 cross-sections

Nb,Rd = χAfy

γM1(6.13a)

and for Class 4 sections

Nb,Rd = χAefffy

γM1(6.13b)

where

χ is the reduction factor for the relevant buckling mode which is generally ‘flexuralbuckling’. In other cases ‘torsional’ or ‘flexural–torsional’ modes may govern.

χ = 1[� + (�2 − λ2)1/2] ≤ 1

� = 0,5[1 + α(λ − 0,2) + λ2]

λ =(

Afy

Ncr

)1/2

= (Lcr/i)λ1

λ1 = π

(Efy

)1/2

= 93,9ε

ε =(

235fy

)1/2

The imperfection factor (α) corresponding to the appropriate buckling curve isobtained from Table 6.1, EN 1993-1-1 (2005).

The value of the cross sectional area is A for Classes 1, 2 and 3 sections with noreductions for local buckling provided that the maximum width-to-thickness ratiosare within the limits of Table 5.2, EN 1993-1-1 (2005). For Class 4 sections the effec-tive area, reduced because of local buckling, is calculated from cl 4, EN 1993-1-5(2003).

EXAMPLE 6.2 Angle strut in a roof truss. A steel roof truss is composed of anglesand tee sections and has been analysed assuming pin joints. A particular memberis subject to the following axial forces. Permanent action (FG = −7,2 kN), variablesnow load (Fs = −10,7 kN), variable wind downwards (Fw = −2,0 kN) and variablewind upwards (Fw = +19,1 kN). If an angle, welded at the ends, is chosen to resist theloads, determine the size if the actual length is 2,1 m.

Design load cases:

(a) no wind

1,35FG + 1,5 Fs = 1,35(−7,2) + 1,5(−10,7) = −25,77 kN (compression)

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(b) wind down

1,35 FG + 1,5(Fs + Fw) = 1,35(−7,2) + 1,5(−10,7 − 2)

= −28,77 kN (compression)

(c) wind up

1,0 FG + 1,5 Fw = −7,2 + 1,5(+19,1) = +21,45 kN (tension)

Design value Nb,Ed = 28,77 kN assuming uniform compressive stress across the sec-tion. Try a 65 × 50 × 6 mm angle grade S275 steel, long leg attached (A = 659 mm2,iv = 10,7 mm from Section Tables).

Design as an axially loaded Class 3 compression member, ignoring eccentric loads(cl BB 1.2, EN 1993-1-1 (2005)). The effective area need not be reduced because ofend holes (cl 6.3.1.1(4), EN 1993-1-1 (2005)).

Check maximum width-to-thickness ratios (Table 5.2, EN 1993-1-1 2005)

ht

= 656

= 10,8 <

[15 ×

(235275

)1/2

= 13,9

]satisfactory

b + h2t

= 50 + 652 × 6

= 9,58 <

[11,5 ×

(235275

)1/2

= 10,6

]

satisfactory, no reduction of area for local buckling.

Non-dimensional slenderness about the v–v axis (Eq. (6.50), EN 1993-1-1 (2005))

λv =(

AfyNcr

)1/2

= (Lcr/iv)λ1

= (Lcr/iv)93,9ε

= 2100/10,7[93,9 × (235/275)1/2]

= 2,26

Largest effective slenderness ratio is about the v–v axis for an angle (cl BB 1.2, EN1993-1-1 (2005))

λeff,v = 0,35 + 0,7λv = 0,35 + 0,7 × 2,26 = 1,93

For λeff,v = 1,93 and curve b (Table 6.2, EN 1993-1-1 (2005)) the buckling reductionfactor (Fig. 6.4, EN 1993-1-1 (2005))

χ = 0,23

Or by calculation (Eq. (6.49), EN 1993-1-1 (2005)) the buckling reduction factor

χ = 1[� + (�2 − λeff,v2)1/2]

= 1[2,66 + (2,662 − 1,932)1/2]

= 0,223 < (graph value 0,23)

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where

� = 0,5[1 + α(λeff,v − 0,2) + λeff,v2]

= 0,5 × [1 + 0,34 × (1,93 − 0,2) + 1,932] = 2,66

which includes the imperfection factor for buckling curve b (Table 6.1, EN 1993-1-1(2005))

α = 0,34

Design buckling resistance (Eq. (6.47), EN 1993-1-1 (2005))

Nb,Rd = χAfy

γM1= 0,223 × 659 × 275

1,1 × 1E3

= 36,7 > (Nb,Fd = 28,77) kN satisfactory.

6.4 COMBINED BENDING AND AXIAL FORCE – WITH BUCKLING(cl 6.3.3, EN 1993-1-1 (2005))

6.4.1 Introduction

In practical situations axial forces in columns are accompanied by bending momentsacting about the major and minor axes of bending. The axial force and bending momentvary along the length of the member and an exact analysis is complicated (Culver,1966). In some situations overall buckling, lateral torsional buckling and local bucklingcan occur together. This is more likely to be a problem with Class 4 sections whereoutstand/thickness ratios are large.

Exact theoretical solutions for buckling of columns are not available and in any casewould be too complicated for design. Alternatively design interaction equations (Eqs(6.61) and (6.62), EN 1993-1-1 (2005)) are used based on elastic and plastic limits.A steel member subject to bending moments and axial forces fails when a stress isgreater than first yield but less than full plasticity of the section. First yield theoriesare conservative while full plasticity theories are unsafe.

Joint rotation reduces the buckling load of a column and this can now be incorporatedinto the analysis of frames (cl 2, EN 1993-1-8 (2005)). This can be important becauseeven small rotations can significantly reduce load capacity.

In design it is often assumed that parts of frames can be analysed separately and thecompression members can be isolated and analysed accordingly. For traditional simpledesign methods beam and column structures are assumed to be connected togetherwith pin joints and braced to prevent sidesway collapse. The pin joints are assumedeccentric to the column axis and thus introduce bending moments to the column whichreduces the failure load.

Ch06-H5060.tex 11/7/2007 14: 10 page 191


3m

3m

4,6

m

120

180

180

12

18

18

D

C

B

y

z y

z

A

FIGURE 6.7 Example: three storey corner column

EXAMPLE 6.3 Bending moments in a three storey corner column. The first threestoreys of a corner stanchion are sketched in Fig. 6.7. The beam support reactionsare indicated in kN on each beam. Assuming simple design, eccentric pin jointsand cross bracing to reduce sway, determine the bending moments in the columnsAB and BC.

Try the following column sizes from Section Tables.

Columns AB and BC: 203 × 203 × 60 UC

h = 209,6 mm, Iy = 6103 cm4, Iz = 2047 cm4.

Column CD: 203 × 203 × 46 UC

h = 203,2 mm, Iy = 4565 cm4, Iz = 1539 cm4.

Consider a corner column buckling with bending about the major and minor axes.

Bending about the major y–y axis

Stiffness of columns (cm units)

Column AB, KAB = Iy/LAB = 6103/460 = 13,27

Column BC, KBC = Iy/LBC = 6103/300 = 20,34

Column CD, KCD = Iy/LCD = 4564/300 = 15,21

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(a) First floor level (at joint B):Moment about the y–y axis from eccentricity of the joint assuming

MBy = (beam reaction)(100 + h/2)

= 180 × (100 + 209,6/2)1E3

= 36,86 kNm.

For simple design moments are distributed in proportion to stiffness

MBA = MByKAB

KAB + KBC= 36,86 × 13,27

13,27 + 20,34= 14,55 kNm

MBC = MByKBC

KAB + KBC= 36,86 × 20,34

13,27 + 20,34= 22,31 kNm

(b) Second floor level (at joint C):Assuming that the splice between columns BC and CD lies above first floorlevel, then the design moment is the same at the first floor (MCy = 36,86 kNm). Ifmoments are distributed equally for simple multi-storey construction

MCB = MCD = MCy

2= 36,86

2= 18,43 kNm.

It should be appreciated that the simple method does not incorporate joint stiffnesswhich reduces moments at mid-span for the beams. This is beneficial, but momentsin the column are inaccurate. However, the method has been used extensively inthe past.

EXAMPLE 6.4 Two storey corner column. Determine the size of a column sectionrequired for a two storey corner column shown in Fig. 6.8. The design loads (kN)shown are the end reactions from the beams. The beams are connected to the columnusing cleats and bolts and pin joints are assumed. The columns and beams are encasedin concrete.

3,7

m4,

6m

117

180

12

18

C

B

y

z y

z

A

FIGURE 6.8 Example: two storey corner column

Ch06-H5060.tex 11/7/2007 14: 10 page 193


Axial design loads on the column:

(a) First floor kNFloor beam (self-weight included) 117Wall beam (self-weight included) 12Self-weight of column BC (estimated) 10Total 139

(b) Ground floor kNFloor beam (self-weight included) 180Wall beam (self-weight included) 18Upper storey 139Total (excluding self-weight of column AB) 337

Try 152 × 152 × 37 kg UC grade S275 steel for both storeys. From SectionTables h = 161,8 mm, b = 154,4 mm, tw = 8,1 mm tf = 11,5 mm, Agross = 4730 mm2,iz = 38,7 mm, iy = 68,4 mm, Wel,y = 274E3 mm3, Wel,z = 91,5E3 mm3, Wpl,y =309E3 mm3, Wpl,z = 140E3 mm3, Iz = 706E4 mm4, Iy = 2213E4 mm4, Iw = 0,0399 dm6,It = 19,3E4 mm4.

This is a Class 1 section, that is, c/t outstand ratios are within limits (Table 5.2, EN1993-1-1 (2005)) and local buckling does not occur. However, the section must bechecked for overall buckling and lateral torsional buckling.

Assuming simple nominal bending moments from eccentricity of the beam reactions(R) at join B:

MBy = R(100 + h/2)1E3

= 180(100 + 161,8/2)1E3

= 32,56 kNm

MBz = R(100 + tw/2)1E3

= 18(100 + 8,1/2)1E3

= 1,87 kNm

Design bending moments applied to the column (same section throughout)

My,Ed = MBy(IBC/LBC)[(IBC/LBC) + (IBA/LBA)]

= 32,56(1/3,7)[(1/3,7) + (1/4,6)]

= 32,56 × 0,554 = 18,03 kNm

Mz,Ed = MBz × 0,554 = 1,87 × 0,554 = 1,036 kNm

Check combined axial load, bending and lateral torsional buckling about the strongery–y axis using the interaction formula (Eq. (6.61), EN 1993-1-1 (2005))

NEd

χyNRk/γM1+ kyy(My,Ed + �My,Ed)

χLTMy,Rk/γM1+ kyz(Mz,Ed + �Mz,Ed)

Mz,Rk/γM1≤ 1

= 337946

+ 0,465 × (18,03 + 0)0,751 × 77,3

+ 0,453 × (1,036 + 0)35

= 0,356 + 0,144 + 0,013 = 0,513 < 1 satisfactory.

Ch06-H5060.tex 11/7/2007 14: 10 page 194


Check combined axial load, bending and lateral torsional buckling about the weakerz–z axis using the interaction formula (Eq. (6.62), EN 1993-1-1 (2005))

NEd

χzNRk/γM1+ kzy(My,Ed + �My,Ed)

χLTMy,Rk/γM1+ kzz(Mz,Ed + �Mz,Ed)

Mz,Rk/γM1≤ 1

= 337532

+ 0,453 × (18,03 + 0)0,751 × 77,3

+ 0,755 × (1,036 + 0)35

= 0,633 + 0,141 + 0,022 = 0,796 < 1 satisfactory.

The two previous equations include the following values.

For buckling load calculations, consider column AB just below B, assuming the buck-ling length about the y–y and z–z axes, Lcr = 0,85 L. Use buckling curve b for bucklingabout the y–y axis and curve c for the z–z axis (Table 6.2, EN 1993-1-1 (2005)) basedon h/b < 1,2; tf < 100 mm and steel grade S275.

The non-dimensional slenderness ratio (Eq. (6.50), EN 1993-1-1 (2005))

λy =(

AfyNcr

)1/2

=(

Lcr

iy

)× 1

λ1

= (0,85 × 4600)68,4

× 1[93,9 × (235/275)1/2]

= 0,659

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve b), λy = 0,659 and a reduction factorχy = 0,80 the design buckling resistance for a Class 1 section (Eq. (6.47), EN 1993-1-1(2005))

Nb,Rd = χyAfy

γM1= 0, 80 × 4730 × 275

1,1 × 1E3= 946 kN.

My,Rd = Wpl,yfy

γM1= 309E3 × 275

1,1 × 1E6= 77, 3 kNm.

λz =(

AfyNcr

)1/2

=(

Lcr

iz

)× 1

λ1

=(

0,85 × 460038,7

)× 1

[93,9 × (235/275)1/2]= 1,16

From Fig. 6.4, EN 1993-1-1 (2005) (buckling curve c) and λz = 1,16 the reductionfactor χz = 0,45 and the design buckling resistance for a Class 1 section (Eq. (6.48),EN 1993-1-1 (2005))

Nb,Rd = χzAfy

γM1= 0,45 × 4730 × 275

1,1 × 1E3= 532 kN.

Mz,Rd = Wpl,zfy

γM1= 140E3 × 275

1,1 × 1E6= 35 kNm.

Ch06-H5060.tex 11/7/2007 14: 10 page 195


For a Class 1 section (Table 6.7, EN 1993-1-1 (2005))

�My,Ed = �Mz,Ed = 0

Interaction factor for a Class 1 section (Table B1, EN 1993-1-1 (2005)). The lesservalue of

kyy = cmy

[1 + (λy − 0,2)NEd

χyNRk/γM1

]

= 0,4 ×[

1 + (0,659 − 0,2) × 337E30,80 × 4730 × 275/1,1

]= 0,465

or

kyy = cmy

[1 + 0,8NEd

χyNRK/γM1

]

= 0,4 ×[

1 + 0,8 × 337E30,80 × 4730 × 275/1,1

]= 0,514 > 0,465 use 0,465

where (Table B3, EN 1993-1-1 (2005))

cmy = 0,6 + 0,4ϕ = 0,6 + 0,4 × (−0,5) = 0,4

Interaction factors for a Class 1 section (Table B1, EN 1993-1-1 (2005)). The lesservalue of

kzz = cmz

[1 + (2λz − 0,6)

NEd

χzNRk/γM1

]

= 0,4 ×[

1 + (2 × 1,16 − 0,6) × 337E30,45 × 4730 × 275/1,1

]= 0,836

or

kzz = cmz

[1 + 1,4NEd

χzNRk/γM1

]

= 0,4 ×[

1 + 1,4 × 337E30,45 × 4730 × 275/1,1

]

= 0,755 < 0,836 use 0,755

where (Table B3, EN 1993-1-1 (2005))

cmz = 0,6 + 0,4ϕ = 0,6 + 0,4 × (−0,5) = 0,4

and

kyz = 0,6kzz = 0,6 × 0,755 = 0,453

Ch06-H5060.tex 11/7/2007 14: 10 page 196


Reduction factor for lateral torsional buckling (Eq. (6.57), EN 1993-1-1 (2005))

χLT = 1

[�LT + (�2LT − βλ

2LT)1/2]

= 1[0,881 + (0,8812 − 0,75 × 0,9052)1/2]

= 0,751 < 1

or

1

λ2LT

= 10,9052 = 1,22 > 1, use 0,751

which includes the factor

�LT = 0,5[1 + αLT(λLT − λLT,0) + βλ2LT]

= 0,5 × [1 + 0,21 × (0,905 − 0,2) + 0,75 × 0,9052] = 0,881

where the imperfection factor αLT = 0,21 (Table 6.3, EN 1993-1-1 (2005)).

λLT =(

WplyfyMcr

)1/2

=(

309E3 × 275103,85E6

)1/2

= 0,905

The elastic critical bending moment Eq. (5.5)

Mcr =(

π2EIz

L2

)[Iw

Iz+ L2GIt

π2EIz

]1/2

=(

π2 × 14834,62

)[0,0399E-6

706E-8+ 4,62 × 15,59

π2 × 1483

]1/2

= 103,85 kNm

where

L = 4,6 m

EIz = 210E6 × 706E-8 = 1483 kNm2

GIt = 80,77E6 × 19,3E-8 = 15,59 kNm2

EIw = 210E6 × 0,0399E-6 = 8,379 kNm4

For further information on elastic critical bending moments see Chapter 5.

Check the self-weight of column BC. Minimum overall dimensions of cased column

H = 161,8 + 100 = 261,8, say 270 mm

B = 154,4 + 100 = 254,4, say 260 mm

Ac = Ag − As = 270 × 260 − 4740 = 65460 mm2.

Ch06-H5060.tex 11/7/2007 14: 10 page 197


Total weight = steel column + concrete casing

= Lρg + LAcρg

= 3,7 × 37 × 9,81/1E3 + 3,7 × 65460 × 2400 × 9,81/1E9

= 7 kN < 10 kN (assumed) satisfactory.

These calculations show that a 152 × 152 × 37 kg UC section grade 275 steel issatisfactory, however a lesser weight of 152 × 152 × 30 kg UC might be suitable.

Calculations are more extensive for Class 4 sections which are reduced in area to allowfor local buckling (cl 5.5.2(2) EN 1993-1-1; and cl 4.3 EN 1993-1-5). Section propertiesare based on the effective cross-sections.

REFERENCES

Beer, H. and Schultz, G. (1970). Theoretical basis for the European column curves. ConstructionMetallique, No. 3.

Bleich, F. (1952). Buckling strength of metal structures. McGraw-Hill.Culver, G.C. (1966). Exact solution for biaxial bending equations. American Society of Civil Engineers,

Str. Div., 92(ST2).Dwight, J.B. (1975). Adaption of the Perry formula to represent the new European steel column curves,

Steel Construction, AISC, 9(1).The background to British Standards for structural steel work. Imperial College, London, and Constrado.ECCS (1976). Manual on stability, introductory report, second international colloquium on stability.

European Convention for Structural Steelwork. Liege.EN 1993-1-1 (2005). General rules and rules for buildings. BSI.EN 1993-1-5 (2003). Plated structural elements. BSI.EN 1993-1-8 (2005). Design of joints. BSI.Euler, L. (1759). Sur la force de collones. Memoires de l’Acadamie de Berlin.Godfrey, G.B. (1962). The allowable stress in axially loaded struts, Structural Engineer, March 1962.Nelson, H.M. (1953). Angles in tension. British Constructional Steelwork Association publication

No. 7.Regan, P.E. and Salter, P.R. (1984). Tests on welded angle tension members, Structural Engineer,

62B(2).Robertson, A. (1925). The strength of struts, Selected Engineering Paper No. 28, Institution of Civil

Engineers.Trahair, N.S. and Bradford, M.S. (1988). The Behaviour and Design of Steel Structures. Chapman and

Hall.Williams, F.W. and Sharp, G. (1990). Simple elastic critical load and effective length calculations for

multi-storey rigid sway frames. Proceedings of the ICE, 90.Wood, R.H. (1974). Effective lengths for columns in multi-storey buildings. Structural Engineer, 52.

Ch07-H5060.tex 12/7/2007 14: 23 page 198

C h a p t e r 7 / Structural Joints(EN 1993-1-8, 2005)

7.1 INTRODUCTION

Structural steel connections, referred to as joints in the Code, are required to ensurecontinuity at the intersection members and foundations. They are also used to formsplices and to construct brackets to support loads. Generally structural steel joints arecomposed of plates, or parts of sections, shop welded in controlled conditions andbolted together on site. Welding can be carried out on site but it needs to be carefullysupervised and is limited because of the expense. The physical appearance of somejoints is shown in Figs 7.22 and 7.23.

Structural joints transmit internal forces and moments in a structure and strength isof major importance. However, the rigidity of joints also needs to be considered. Alljoints are semi-rigid with associated small linear and larger rotational movements. Thelinear movements at a joint are generally small and generally need not be considered,but the rotational movements affect the distribution of forces and moments whichmust be taken into account in structural analysis.

For theoretical purposes in the analysis of structures, joints are classified (Table 5.1,EN 1993-1-8 (2005)) by strength and rigidity as:

(a) Pinned – low moment of resistance.(b) Rigid – full strength and all deformations are insignificant.(c) Semi-rigid – characteristics of the connection lie between (a) and (b).

These theoretical and practical descriptions are important to recognize when analyzinga structure to determine the distribution of forces and moments using global analysis(cl 5.1, EN 1993-1-8 (2005)). There are three methods of global analysis:

(1) Elastic – joints are classified by rotational stiffness.(2) Rigid-plastic – joints are classified by strength.(3) Elastic plastic – joints are classified by stiffness and strength.

Ch07-H5060.tex 12/7/2007 14: 23 page 199


7.2 THE IDEAL STRUCTURAL JOINT

The types of joints in common use have been developed and modified to suit themanufacturing and assembly processes and the ideal requirements are:

(a) Simple to manufacture and assemble.(b) Standardized for situations where the dimensions and loads are similar thus

avoiding a multiplicity of dimensions, plate thicknesses, weld sizes and bolts.(c) Manufactured from materials and components that are readily available.(d) Designed and detailed so that work is from the top of the joint not from below

where the workman’s arms will be above his head. There should also be sufficientroom to locate a spanner, or space to weld if required.

(e) Designed so that the welding is confined to the workshops to ensure a good qualityand to reduce costs.

(f) Detailed to allow sufficient clearance and adjustment to accommodate the lackof accuracy in site dimensions.

(g) Designed to withstand normal working loads and also erection forces.(h) Designed to avoid the use of temporary supports during erection.(i) Designed to develop the required load–deformation characteristics at service and

ultimate loads.(j) Detailed to resist corrosion and to be of reasonable appearance.

(k) Low in cost and cheap to maintain.

7.3 WELDED JOINTS

Welding is a method of connecting components by heating the materials to a suitabletemperature so that fusion occurs. The most common method for heating steelworkis by means of an electric are between a coated wire electrode and the materialsbeing joined. The electrical circuit is shown in Fig. 7.1(a). During the process, whichis illustrated in Fig. 7.1(b), the coated electrode is consumed, the wire becomes thefiller material and the coating is converted partly into a shielding gas, partly into slag,and some part is absorbed by the weld metal. This method, known as the manualmetal arc welding process, is still the most common for structural joints because of lowcapital cost and flexibility. However, for long continuous welds automatic processesare preferred because of consistency.

Generally the electrode is stronger than the parent metal. For manual metal arc weld-ing the electrodes should be compatible with the steel being welded (BSEN 499, 1995;Gourd, 1980). The main reason for the flux covering to the electrode in the manualmetal arc welding process is to provide an inert gas which shields the molten metal fromatmospheric contamination. In addition the flux forms a slag to protect the weld untilit is cooled to room temperature, when the slag should be easily detachable. Otherfunctions of the flux include: arc stabilization, control of surface profile, control ofweld metal composition, alloying and deoxidization. However, it should be noted that

Ch07-H5060.tex 12/7/2007 14: 23 page 200

200 • Chapter 7 / Structural Joints (EN 1993-1-8, 2005)

(a) Arc welding circuit (b) Shielded arc welding

CircuitComponentsbeing welded

Welding machine

Electrode

ArcArc stream

Coating onelectrode

Electrode wire

Molten arc pool

GaseousshieldSlag

Weld

Base metal

FIGURE 7.1 Shielded metal are welding

the flux can be a source of hydrogen contamination from absorbed and chemicallycombined moisture. The absorbed moisture can be removed by drying.

The particular advantage of welding is that it forms a rigid joint, however the manu-facture of welded joints requires more skill and supervision than bolted joints. Mostwelded structural joints are affected using the manual metal arc process but long con-tinuous welds, which occur in built-up girders, are laid down with automatic weldingequipment. The automatic processes achieve the exclusion of atmospheric pollutionby gas shielding, flux core or submerged arc.

Types of welds used in structural engineering and allowed in the European Code arefillet, slot, butt, plug and flare groove. Some common types are shown in Fig. 7.2.

The two types in most common use are butt and fillet welds. Butt welds, oftenused to lengthen plates in the end on position, may be considered as strong as theparent plate as long as full penetration for the weld is achieved. For thin platespenetration is achieved without preparing the plate, but on thicker plates V or dou-ble J preparation is required. Butt welds are also used to connect plates at rightangles but the plates require edge preparation. Partial penetration butt welds are notfavoured in design and should not be used intermittently or in fatigue situations (BSEN1011-1 (1998)).

Fillet welds are generally formed with equal leg lengths. They do not require specialedge preparation of the plates and are therefore cheaper than butt welds.

Generally in connections, plates intersect at right angles but intersection angles ofbetween 60◦ and 120◦ can be used provided that the correct throat size is used indesign calculations. In order to accommodate lack of fit the minimum leg length offillet weld in structural engineering is 5 mm although 6 mm is often preferred. Themaximum size of fillet weld from a single run metal arc process is 8 mm, but 6 mm is

Ch07-H5060.tex 12/7/2007 14: 23 page 201


Butt weld madefrom one side

Butt weld madefrom two side

T-butt weld madefrom two sides

T-butt weld madefrom two sides

t

c

db

b � d t andc 3 mm or c t /5

Slope should not exceed1:1 in butt joints

T-butt weldThe weld is regarded to be of equal orhigher strength than the vertical member

T-fillet weld Lap joint

(b) Examples of fillet welds

(a) Examples of butt welds

FIGURE 7.2 Types of welds in structural joints

preferred to guarantee quality. When larger fillet welds are required they are formedfrom multiple runs.

The use of intermittent butt and fillet welds is permitted (cl 4.3.2.2, EN 1993-1-8(2005)). Intermittent welds are not favoured in structural engineering because theyintroduce stress discontinuities, act as stress raisers, may introduce fatigue cracks,may act as corrosion pockets and are difficult to produce with an automatic weldingmachine. The spacing of intermittent fillet welds is shown in Fig. 4.1, EN 1993-1-8(2005).

Ch07-H5060.tex 12/7/2007 14: 23 page 202


a

a

a

aa

Deeppenetration

FIGURE 7.3 Throat thickness of fillet welds (Figs 4.3 and 4.4, EN 1993-1-8)

7.3.1 Throat Thickness of a Weld (cl 4.5.2, EN 1993-1-8 (2005))

The size of a weld is often described by the leg length but the strength is calculatedusing the effective throat thickness (a) as defined in Fig. 7.3. The effective throatthickness should not be less than 3 mm.

7.3.2 Effective Length of a Weld (cl 4.5.1, EN 1993-1-8 (2005))

The effective length of a fillet weld should be taken as the length over which the fillet isfull size. The minimum length allowed to transmit loading is six times the throat thick-ness, and not less than 30 mm. In practice, fillet welds terminating at the ends, or sides,of parts are returned continuously round the corners for a distance of not less than twicethe leg length, unless impracticable. The continuation round the corner is to reducestress concentrations and its strength is generally ignored in strength calculations.

The effective length of a weld is reduced if a component distorts under load in situ-ations similar to that shown In Fig. 7.4, where the deformations in the weld adjacentto the web are greater than those at the end of the flange. The larger deformations atthe web initiate failure in the weld at this point with consequent loss of strength forthe total length of the weld (Elzen, 1966; Rolloos, 1969).

For design the effective breadth (beff ) of a weld (cl 4.10, EN 1993-1-8 (2005)) is: Fora rolled ‘I’ or ‘H’ section (Eqs (4.6a) and (4.6b), EN 1993-1-8 (2005))

beff = tw + 2r + 7ktf ≤ tw + 2r + 7(

tftp

)(fy,f

fy,p

)tf (7.1a)

For box or channel sections where the widths of the connected plate and the flangeare similar

beff = 2tw + 5 tf ≤ 2tw + 5(

tftp

)(fy,f

fy,p

)tf (7.1b)

7.3.3 Long Welded Joints (cl 4.11, EN 1993-1-8 (2005))

The stress distribution along the length of a long lap joint is not uniform being greatestat the ends. To allow for this the length of the weld is reduced. For joints longer

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Deformation of weldsdue to shear force Vdetermining flexiblebeam flanges

V

Bendingmoment M

Bracket I section

ColumnI section

Side elevation End elevation

Shearforce V

Deformation of welddue to bending momentM deforming theflexible column flange

Plan

FIGURE 7.4 Reduction in strength of welds associated with flange deformations

than 150 times the throat thickness (a), the reduction factor (Eq. (4.9), EN 1993-1-8(2005))

βLW,1 = 1, 2 − 0, 2Lj/(150a) ≤ 1 (7.2)

where

Lj is the overall length of the lap in the direction of the force transfer.

7.3.4 Design Resistance of Fillet Welds (cl 4.5.3,EN 1993-1-8 (2005))

The real external forces acting on a 90◦ fillet weld are probably those shown inFig. 7.5 (a) (Clarke, 1971). Experiments (Biggs et al., 1981) on 90◦ fillet welds ofequal leg length loaded to failure show that the fracture plane varies between 10◦ and80◦ depending on the combination of external forces. The actual distribution of stresson the failure plane is uncertain but a theoretical distribution (Kato and Morita, 1974)shows peak stresses at the root of the weld which reduce towards the face of the weld.This distribution appears to be confirmed by experimental observations of cracks initi-ating at the root. The situation is complicated further by residual stresses and variablessuch as the type of electrode type of steel, ratio of the size of weld to the plate thick-ness, the quality of weld and whether the loading is static or dynamic. If stresses onthe failure plane are assumed to be uniform then the relationship between the averageshear stress and tensile stress on the failure plane has been shown (Biggs et al., 1981)to approximate to an ellipse. An ellipse of failure stresses combined with a variable

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Mwz

Fwz

Fwy

Fwz

Fwy

Fwx

FwzFwx

Fwy

Mwy

Mwx

Fwx

zx

yf

f = 45°

f = 45°

f sf

s

τf

τ1

τ2

(a) Complex system of forcesLe

g le

ngth

s

Unit length

Throa

t

thick

ess a

(b) Simple system of forces

FIGURE 7.5 Forces acting on a 90◦ fillet weld

fracture angle can be used theoretically to predict the magnitude of the external forces,but the method is unnecessarily complicated for design purposes.

For design purposes a complex system of external forces acting on a fillet weld isreduced forces acting in three perpendicular directions on a unit length of weld asshown in Fig. 7.5 (b). The vector sum of all the design forces should not exceedthe design resistance of a fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005)) and may beexpressed as

F2wx + F2

wy + F2wz ≤ F2

w,Rd (7.3)

The term Fw,Rd = a fvw,d = a( fu/31/2)/(βw γM2) is the design strength of a fillet weldper unit length.

fu = nominal ultimate tensile strength of the weaker part joined (Table 3.1, EN 1993-1-1(2005))

βw = correlation factor dependent on steel grade (Table 4.1, EN 1993-1-8 (2005))

γM2 = partial safety factors for joints, recommended value 1,25 (Table 2.1, EN 1993-1-8(2005)).

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The design strength of a fillet weld has been shown (Ligtenberg, 1968) to be related tothe strength of the parent material. The correct type and strength of electrode mustbe used for each grade of steel.

An alternative directional method of design (cl 4.5.3.2, EN 1993-1-8 (2005)) involvescalculating the normal and shear stresses on the throat section of the weld and com-bining them using the yield criteria developed in Chapter 2. This method is morelaborious and introduces the possibility of further errors when resolving the forcesonto the critical plane. The relationship between the forces Fx, Fy and Fz for thismethod can be expressed I the same form as Eq. (7.3) for a 45◦ plane (Holmes andMartin, 1983). The size of the fillet weld obtained using this method is slightly lessthan using the vector addition method.

7.3.5 Load–Deformation Relationships for Fillet Welds

The strength of the weld in a connection is of primary importance but the load–deformation characteristics of the weld should also be considered. The deformationat the maximum load varies from approximately 0.6 to 1.4 mm depending on the ori-entation of the weld in relation to the applied load (Clarke, 1970) as shown in Fig. 7.6.The maximum deformation for the side fillet weld which is parallel to the applied loadand the minimum is for an end fillet weld. To allow for this effect design stresses arebased on the weaker side fillet welds. At the stress the disparity in deformations forend and side fillet welds is less than at failure.

7.3.6 Conditions Affecting the Strength of Welded Joints

The following conditions affect the strength of welded connections:

(a) Use of an incorrect steel (BS 7668, 1994).(b) Use of an incorrect electrode (BSEN 499, 1995).(c) Cavities and slag inclusions in the weld. These may be detected by non-destructive

testing.

Design stress

Test weld

P

P

θ varies

210

Weld deformation (mm)

Side fillet weld

Ave

rage

thro

at s

tres

s (M

Pa)

500

End fillet weld

θ = 0°

θ = 30°

θ = 60°θ = 90°

FIGURE 7.6 Load-deformationrelationships for an 8 mm filletweld (Clarke 1970)

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(d) Excessive lack of fit between components.(e) Stress concentrations combined with oscillating loads producing fatigue.(f) Residual stresses introduced from differential heating during welding.(g) Hydrogen cracks associated with welding occur when the cooling rate is too rapid

(Fig. 7.7(a)). Excessive hardening occurs in the heat affected zone which cracksunder the action of residual stresses if sufficient hydrogen is present in the weld.This defect can be avoided by controlling the cooling and the hydrogen input tothe weld (BSEN 1011-1 (1998)).

(h) Lamellar tearing may occur when welding plate connections of the type shown inFig. 7.7 (b). the cracks are produced by a combination of low ductility in the plate inthe transverse direction and high point restraint in the weld which induces tensileforces adjacent to the connection. The low ductility in the plate is produced byinclusions of non-metallic substances formed in the steel making process. Whenthe ingot is rolled to make steel these inclusions form as plates parallel to thedirection of rolling. Only a small percentage of plates are susceptible to lamellartearing, and where it occurs joint details can be changed to reduce the chances ofif affecting the strength of the connection (Farrar and Dolby, 1972).

(i) Brittle fracture.(j) Corrosion which reduces the size of components or causes pitting which may

initiate fatigue cracks.(k) Insufficient penetration of the parent metal which leads to a reduction in strength

of the weld. The welder uses a voltage and arc length which produces a stablearc and a satisfactory weld profile. The current then becomes the main factor incontrolling penetration. Another important factor in depth of penetration is edgepreparation. Plates of 6 mm with square edges can be butt welded from one side,but the edges of thicker material must be bevelled to provide access for the arc.

(l) Lack of side wall fusion occurs if there is poor bond between the parent and weldmetal. Good bonding can only occur when the surface of the parent metal hasbeen melted before the weld metal is allowed to flow into the joint.

Further information on faults in welds can be found in Gourd (1980).

Fillet weld Fillet weld

Hydrogen cracks

(a) Hydrogen cracks (b) Lamellar tearing

Lamellar tearing

FIGURE 7.7 Faults associated with welding

Ch07-H5060.tex 12/7/2007 14: 23 page 207


7.3.7 Design Strength of Fillet Welds (cl 4.5.3.3 and Table 4.1,EN 1993-1-8 (2005))

The design strength of a unit length of fillet weld (Eq. (4.4), EN 1993-1-8 (2005))

Fw,Rd = (throat thickness) × (unit length) × (design shear stress)/factors

=a(

fu31/2

)βwγM2

(7.4)

The factor (βw) is related to steel grade, and varies between 0,8 and 1,0 (Table 4.1,EN 1993-1-8 (2005)). The ultimate stress ( fu) is that of the weaker material joined(Table 3.1, EN 1993-1-1 (2005)). Typical values are fu = 430 MPa for S275 grade steel,and fu = 510 MPa for S355 grade steel. The value ( fu/31/2) is from the shear distortionstrain energy theory as explained in Chapter 2. It should be noted that although thestrength of the weld is calculated using the throat thickness (a) (Figs 4.3 and 4.4, EN1993-1-8 (2005)) the weld is often specified by the leg length. A table of the strengthof fillet welds is given in Annex Al.

7.4 BOLTED JOINTS (cl 3, EN 1993-1-8 (2005))The advantage of bolted joints is that they require less supervision than welding, andtherefore are ideal for site conditions. Other advantages are that the connection can befastened quickly, supports load as soon as the bolts are tightened and accommodatesminor discrepancies in dimensions.

Disadvantages of bolted connections are that for large forces the space required forthe joint is extensive, and the connection is not as rigid as a welded connection evenwhen friction grip bolts are used.

Steel bolts are identified by their gross diameter, strength and use. The preferred sizesof bolts in general use are 16, 20, 24, 30 and 36 mm diameter. The most common sizeuse in structural connections is 20 mm.

The types of bolt in common use (Table 3.1, EN 1993-1-8 (2005)) are:

(a) Ordinary bolts Classes 4.6–6.8 and includes foundation bolts.(b) Pre-loaded bolts Classes 8.8 and 10.9.

A Class 4.6 bolt is low in cost, can be installed with the use of simple tools andrequires little supervision during the erection. At fracture the bolt has a relativelylarge extension of 25%, a property which is preferred at plastic collapse.

Where forces are large, or where space for the connection is limited, or where erectioncosts can be reduced by using fewer bolts, then the higher grade bolts are used. The per-centage elongation of 12% at failure is less, but is still acceptable for design purposes.

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Where a more rigid bolted connection is required, for example in plastic methodsof design, pre-loaded bolts are used. The strength of these bolts is greater with anincreased cost for the additional site supervision which is necessary to ensure that thebolts are axially pre-loaded in tension to the design values. The object of the pre-loadis to ensure that the friction between the ‘faying’ surfaces prevents slip when subjectto external shear forces and thus produce a more rigid joint.

The nuts of pre-loaded bolts are tightened with a torque wrench which is calibrated inrelation to the required axial pre-load. A simpler method of measuring the axial forcein the bolts is to use a load indicating washer, under the head of the nut, which reducesin thickness to a specified value for a specified pre-load. The washer is less accuratethan the torque spanner (Bahia and Martin, 1981). A further alternative method ofensuring that the bolt is pre-loaded is to specify ‘turns of the nut’. Investigations (Fisherand Struck, 1974) showed that in general the pre-load produced by the torque wrenchand ‘turns of the nut’ method on site exceeded the specified value.

Close tolerance turned bolts are used only where accurate alignment of componentsor structural elements is required. The shank of the bolt is at least 2 mm greater indiameter than the threaded portion of the bolt and the hole is only 0,15 mm greaterthan the shank diameter. This small tolerance necessitates the use of special methodsto ensure that the holes align correctly.

Foundation bolts, or holding down bolts, are used for connection structural elementsto concrete pads or concrete foundations. Generally the bolts are cast into the concretebefore erection of the steel work and thus require accurate setting out. Where upliftforces occur the bolts must be anchored by a washer plate. Most bolts used are Class4.6 but higher strengths are available. Sometimes bolts are grouted in the holes duringerection using epoxy resin.

Rivets were used extensively in the past in the fabrication shop and on site. They weredifficult and expensive to place but they resulted in a rigid connection because thehot rivet, after driving, expanded to fill the hole. Rivets have now been superseded bywelding and bolts.

There are five categories of bolted joints (Table 3.2, EN 1993-1-8 (2005)) related tothe type of connection or bolt and pre-loading.

Shear connections

Category A: bearing type where there is no pre-loading nor special provision for con-tact surfaces. Design for shear and bearing resistance. This is the cheapest type ofconnection where complete rigidity and plasticity are not important.

Category B: slip resistant at serviceability limit state. Design for slip resistance atthe serviceability limit state and shear and bearing resistance at the ultimate limitstate. Connection used to provide full rigidity in the elastic stage of behaviour whendeflections are critical.

Ch07-H5060.tex 12/7/2007 14: 23 page 209


Category C: slip resistant at ultimate limit state. Design for slip resistance and bearingresistance at the ultimate limit state. Connection used to limit movements at theultimate plastic limit state.

Tension connections

Category D: No pre-loading of bolts.

Category E: Pre-loaded bolts.

7.4.1 Washers

In British practice most bolts have steel washers under the head and under the nut. Thewasher distributes the bolt force and prevents the nut, or bolt head, from damagingthe component or member. However washers are not essential in all cases (ECSS,1981), and they are now being omitted in British practice.

Ordinary washers (BSEN 4320, (1998)) are in general use but hardened washers(BSEN 14399-1 to 5 (2005)) are used for pre-loaded bolts. The outside diameter of awasher is an important dimension when detailing, for example to avoid overlappingan adjacent weld.

7.4.2 Bolt Holes

Bolt holes are usually drilled, but may be punched full size, or punched under size andreamed. Holes should never be formed by gas cutting because of the inaccuracy andthe effect on the local properties of the steel.

Punched holes are preferred by steel fabricators because it saves time and reducescost. However research (Owens et al., 1981) shows that distortion in the vicinity of ahole reduces toughness and ductility and can lead to brittle fracture. Punched holesshould not be used in locations where plastic tensile straining can occur.

Bolt holes are made larger than the bolt diameter to facilitate erection and to allow forinaccuracies. The clearance is 2 mm for bolts not exceeding 24 mm diameter and 3 mmfor bolts exceeding 24 mm diameter. Oversize and slotted holes are allowed but notoften used. Slotted holes are sometimes used for pre-loaded bolts to facilitate erectionwith unusual shaped structures, or alternatively they can be used to accommodatemovement in a structure. The clearance for a close tolerance turned bolt is 0,15 mm.

Bolt holes reduce the gross cross-sectional area of a plate to the net cross-sectionalarea. The net value is used for calculations where the structural element, or parts of anelement, are in tension. Bolt holes also produce stress concentrations, but it is arguedthat these are offset by the fact that at yield the highly stressed cross-section will workharder before fracture and yield will by then have occurred at adjacent cross-sections.The gross cross-section of a member is used in compression because at yield the bolt

Ch07-H5060.tex 12/7/2007 14: 23 page 210


hole deforms and the shank of the bolt resists part of the load in bearing. Considerationshould be given to corrosion and local buckling when deciding the position of holes.

7.4.3 Spacing of Bolt Holes (Table 3.3, EN 1993-1-8 (2005))

The longitudinal spacing between the centre line of bolts in the direction of the axialstress in a member is called the pitch. The minimum spacing in the direction of theload of 2,2 × (diameter of the hole) is specified to prevent excessive reduction in thecross-sectional area of a member, to provide sufficient space to tighten the bolts andto prevent overlapping of the washers. Other critical distances are given in Table 3.3and Fig. 3.1, EN 1993-1-8 (2005). These values are specified to prevent buckling ofplates in compression between bolts, to ensure that bolts act together as a group toresist forces, and to minimize corrosion.

7.4.4 Edge and End Distances for Bolt Holes (Table 3.3, EN1993-1-8 (2005))

Edge and end distances are specified to resist the load, to prevent local buckling, tolimit corrosion and to provide space for the bolt head, washer and nut.

The edge distance is from the centre of a hole to the nearest edge measured atright angles to the direction of the load. The minimum edge distance specified is1,2 × (diameter of the hole) and the maximum should not exceed 4t + 40 mm.

The end distance is from the centre of a hole to the adjacent edge in the direction of theload transfer. The minimum end distance in the direction of the load is 1,2 × (diameterof the hole) and the maximum should not exceed 4t + 40 mm. The end distance shouldalso be sufficient for bearing capacity.

There are recommended positions, spacing and diameter of holes in Section Tables.These distances are based on providing sufficent clearances to the web and adequateedge distances (Annexes A4 to A6).

7.4.5 Deductions for Holes in Tension Members (cl 3.10, EN1993-1-8 (2005))

Holes are drilled in tension members to accommodate fasteners at connections. A holereduces the gross cross-sectional area and weakens a tension member because thefastener in the hole does not transmit the axial force. In contrast a hole in a compressionmember has little effect on the buckling strength because as the member compressesthe axial force is transmitted by bearing on the shank of the bolt.

When designing a tie the net cross-sectional area is used in calculations to determinethe design axial force. The net cross-sectional area is the gross area reduced by themaximum sum of the sectional areas of the holes. These holes may be in line at right

Ch07-H5060.tex 12/7/2007 14: 23 page 211


A

A B C s = 95

B C

95 95 95 95 95

110

110

430

P =

110

Directionof stress

FIGURE 7.8 Example: net area of a plate

angles to the axial stress in the member (line AA in Fig. 7.8), or staggered (lines BBand CC in Fig. 7.8).

Typical areas to be deducted for bolt holes are:

Bolt diameter + 2 mm clearance for bolts not exceeding 24 mm in diameter or Boltdiameter + 3 mm clearance for bolts exceeding 24 mm in diameter.

For staggered fastener holes the area to be deducted shall be the greater of:

(a) the deduction for non-staggered holes;(b) the sum of the sectional areas of all holes in any diagonal or zig-zag line extending

progressively across the member, or part of the member, less s2t/4p for each gaugespace in the chain of holes.

where (Fig. 7.8)

s is the staggered pitch

p is the spacing of the holes

t is the thickness of the holed material.

For sections such as angles with holes in both legs the spacing is approximately the sumof the back marks to each hole, less the leg thickness. The arrangement and spacingof holes in a member should not significantly weaken a member at a section.

EXAMPLE 7.1 Net area of a plate with holes. Calculate the net cross-sectional areafor the plate shown in Fig. 7.8 which is subject to a tensile force. The plate is 20 mmthick and contains four lines of staggered holes drilled for 24 mm diameter bolts.

From Fig. 7.8, s = 95 mm and p = 110 mm.

Diameter of hole do = d + 2 = 24 + 2 = 26 mm.

Ch07-H5060.tex 12/7/2007 14: 23 page 212


Gross cross-sectional area perpendicular to the direction of stress = 20 × 430 =8600 mm2

Areas to be deducted at possible failure lines are:

nht do − ngss2t4p

where ngs is the number of gauge spaces in the chain of holes.

Line AA: 2 × 20 × 26 = 1040 mm2

Line BB: 3 × 20 × 26 − 1 ×952 × 20/(4 × 110) = 1150 mm2

Line CC: 4 × 20 × 26 − 3 × 952 × 20/(4 × 110) = 849 mm2.

Minimum net area for line BB = 8600 − 1150 = 7450 mm2.

7.4.6 Design Resistance of Single Bolts (Tables 3.1 and 3.4, EN1993-1-8 (2005))

Bolted connections consist of two or more bolts and each bolt may be subject to anycombination of tension, shear or bearing forces. The design resistance of a single boltwhen subject to these forces is now considered in detail.

A bolt in tension fails at the smallest cross-section, that is, the root of the threadswhere the net area is approximately 80% of the gross area. Where the bolt fails acrossthe reduced cross-section at the root of the thread the tension resistance

Ft,Rd = k2fubAs

γM2(7.5)

where the ultimate tensile strength of the bolt (fub) is obtained from Table 3.1. EN1993-1-8 (2005) and k2 = 0,9 except for a countersunk bolt where k2 = 0,63.

Where the bolt assembly fails by the bolt head, or nut, shear punching through theplate then the tension resistance of the plate (Table 3.4, EN 1993-1-8 (2005))

Bp,Rd = 0,6 π dmtpfu

γM2(7.6)

where

fu is the ultimate tensile stress of the plate.

tp is the thickness of the plate under the head of the bolt.

dm is the mean of across points and across flats dimensions of the bolt head or the nut,whichever is smaller.

The shear resistance per shear plane for a single bolt (Table 3.4, EN 1993-1-8 (2005))

Fv,Rd = αvfubA

γM2(7.7)

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Where the shear plane passes through the unthreaded portion of the bolt αv = 0,6 andA is the gross area.

Where the shear plane passes through the threads αv = 0,5 or 0,6, depending on thegrade of the bolt, and A = As the reduced area.

The values of the reduced bolt areas used in this chapter (BS 3692, 2001; BS 4190,2001) are:

Bolt diameter (mm) 12 16 20 22 24 27 30 36Reduced area (mm2) 84,3 157 245 303 353 459 561 817

Values may also be obtained from Section Tables.

Shearing of a bolt occurs on the shank, that is, the gross area of the bolt, if the threadlength on a bolt is carefully specified, but it is safer to assume that it occurs on thereduced area. It also simplifies calculations and avoids confusion. Experiments (Bahiaand Martin, 1980) and other investigators have found shear values that vary between0,62fu and 0,71fu. The value from the Huber–Von–Mises–Hencky shear distortionstrain energy theory is fu/31/2 as shown in Chapter 2.

Ordinary bolts deform when subject to shear stresses but it is important to realize thatthe shear deformation of the connection is increased by the bearing stresses on theplate. The higher the bearing stresses the greater the deformation as shown in Fig. 7.9.

7.4.7 Design Resistance of a Bolt Subject to Shear and TensionForces (Table 3.4, EN 1993-1-8 (2005))

Shear and tensile resistances are related by the linear interaction formula

Fv,Ed

Fv,Rd+ Ft,Ed

1,4Ft,Rd≤ 1, 0 (7.8)

Generally it is assumed that the failure plane passes through the threaded portion ofthe bolt.

This equation is to be compared with a non-linear experimental relationship (Chessonet al., 1965) based on the net cross-sectional area of the bolt (Fig. 7.10). Alternativeelliptical relationships are given in ECSS (1981) and BS 5400 (2000).

7.4.8 Design Bearing Resistance for a Bolt (Table 3.4, EN1993-1-8 (2005))

A bolt subject to a shear force, such as shown in Fig. 7.11, comes in contact with theplate when the shear load is applied and slip occurs. The bearing stresses between thebolt and plate need to be controlled to limit deformations.

Ch07-H5060.tex 12/7/2007 14: 23 page 214


0 2

20

40

60

80

100

120

140

160

40 2 4 6

6 8 10

Total deformation (mm)

High-strength boltsblack bolts

High-strength bolts20 mm diameter

She

ar lo

ad (

kN)

Notes : Permanent deformation of holes shown as a broken line; bolts failed in single shear across the threads; tp is the plate thickness.

Black bolt20 mm diameter

t p =

20

mm

t p =

12,

07m

m

t p =

14,

10m

m

t p =

17,

10m

m

t p =

20,

15m

m

FIGURE 7.9 Relationbetween shear loadand deformation forsingle bolt tests (Bahiaand Martin, 1980)

0 0,2

0,2

0,4

0,6

0,8

1,0

0,4 0,6 0,8 1,0 1,2f t /f tu

f v/f

tu

ft fvftu

⎤⎥⎦

⎤⎥⎦

⎤⎥⎦

⎤⎥⎦

2 2

0,62 ftu+ = 1

Experimental results by Chesson et al., 1965 for highstrength bolts with shear plane onshank and on threads

FIGURE 7.10 Relationshipbetween shear and tensilestresses for bolts

Ch07-H5060.tex 12/7/2007 14: 23 page 215


Bolts

A ASide elevation

d h

Thickness tp

cp

Plan section AA

(a) End bearing failure for a small end distance

(b) Enclosed bearing failure for a large end distance

Plan

Sectional elevation

(c) Actual distribution of bearing stress

(d) Theoretical distribution of bearing stress

FIGURE 7.11 Bearingstresses for bolts

The bolt or plate may deform because of high local bearing stresses between bolt andplate, or a bolt may shear through the end of the plate. The bearing resistance

Fb,Rd = k1αbfudt

γM2(7.9)

Values of k1 control end distances and values of αb control edge distances (Table 3.4,EN 1993-1-8 (2005)). The value of fu is the weaker of the bolt grade or the adjacentplate or section.

The values of the ultimate tensile strength for bolts (Table 3.1 EN 1993-1-8 (2005))used in this chapter are:

Bolt class 4.6 4.8 5.6 5.8 6.8 8.8 10.9fub MPa 400 400 500 500 600 800 1000αv 0,6 0,5 0,6 0,5 0,5 0,6 0,5

Values of steel grades used in examples in this chapter are S275 ( fu = 430 MPa) andS355 ( fu = 510 MPa).

Equation (7.9) assumes uniform bearing stresses as shown in Fig. 7.11(d) whereas inreality they are closer to those shown in Fig. 7.11(c). Design bearing stresses are high

Ch07-H5060.tex 12/7/2007 14: 23 page 216


Total shear deformation (mm)

Permanent deformation of holeshown as a broken line

tp = 12,07 mm

tp = 14,10 mm

tp = 17,10 mm

tp = 20,15 mm

Yield stress of plates221,8 ± 8,03 MPa

0

100

200

300

400

500

600

700

2 4 6 8 10

Ave

rage

bea

ring

stre

ss (

MP

a)

Note: Bolts failed on the threads in single shear at a shear stress of 600 MPa

FIGURE 7.12 Relationships betweenbearing stress and deformation foran M20 high stress single bolt(Bahia and Martin, 1980)

in relation to the yield stress because material subject to bearing stresses is generallyconfined by other parts which restricts deformation. High bearing stresses are not dis-astrous but lead to excessive deformation of a connection as shown in the experimentalresults (Fig. 7.12).

Equations (7.5)–(7.9) express the design tensile and shear strengths of a bolt and canbe presented in the form of tables to reduce calculations (Annex A2).

7.4.9 Bolts Through Packings (cl 3.6.1(12), EN 1993-1-8 (2005))

Where the total thickness of the packing (tp) is greater than three times the nomi-nal diameter (d) of the bolts the design shear resistance is multiplied by a reductionfactor

βp = 9d8d + 3tp

≤ 1 (7.10)

The reason for the reduction in strength is because as the grip length increases thebolt is subject to greater bending moments from shear forces which move furtherapart.

7.4.10 Long Bolt Joints (cl 3.8, EN 1993-1-8(2005))

For long joints the load is not shared equally by the bolts or rivets. The fasteners onthe end resist the greatest force and the resistance gradually reduces to the centre lineof the joint. The reduction is due to friction, errors in marking out and deformations

Ch07-H5060.tex 12/7/2007 14: 23 page 217


Cover plateLj

FIGURE 7.13 Length (Lj) for long bolted joints

in the materials. Where the length of the joint Lj > 15d (Fig. 7.13) measured in thedirection off the transfer force then the design shear resistance reduction factor

βLf = 1 − Lj − 15d200d

(7.11)

but 0,75 ≤ βLf ≤ 1,0

7.4.11 Design of Slip Resistant Joints (cl 3.9,EN 1993-1-8 (2005))

High strength bolts can be pre-loaded when installed and designed to be slip resistantat working load or ultimate load. The design slip resistance of a pre-loaded bolt whensubject to external tensile and shear forces is now considered in detail.

The design slip resistance of pre-loaded high strength bolt of Classes 8.8 or 10.9(cl 3.9.1, EN 1993-1-8 (2005)) is

Fs,Rd = ks n μFp,C

γM3(7.12)

where

Fp,C = 0,7fub As is the design preloading force.

μ = slip factor as listed in Table 3.7, EN 1993-1-8 (2005).

ks = 1 for a bolt in a clearance hole and reduced for slotted holes (Table 3.6, EN1993-1-8 (2005))

n = number of friction interfaces.

Typical Surface treatment Slip factor (μ)

A – blasted with shot or grit, loose rust removed, no pitting 0,50B – ditto, painted with zinc 0,40C – cleaned and loose rust removed 0,30D – surfaces not treated 0,20

Table 7.2 Slip factors(μ) for pre-loaded bolts (Table 3.7, EN 1993-1-8 (2005)).

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The effect of a tensile force acting on a pre-loaded bolt is to reduce the frictionalresistance (cl 3.9.2, EN 1993-1-8 (2005)). The design slip resistance per bolt subject toa combination of shear and tension forces is

Category B connection (Eq. (3.8a), EN 1993-1-8 (2005)) at service load

Fs,Rd,ser = ksnμ(Fp,C − 0,8Ft,Ed,ser)γM3,serv

(7.13)

Category C connection (Eq. (3.8b), EN 1993-1-8 (2005)) at ultimate load

Fs,Rd = ksnμ(Fp,C − 0,8Ft,Ed)γM3

(7.14)

The factor of 0,8 is introduced to allow for the fact that the minimum shank tensionmay not be achieved.

The design tensile, shear and bearing strength of a parallel shank pre-loaded bolt andcan be presented in the form of a table to reduce calculations (Annex A3).

7.5 PLATE THICKNESSES FOR JOINT COMPONENTS

Plates, or parts of sections acting as flange plates, often form part of structural connec-tions. The length and breadth of plates are generally determined from the geometryof the connection but the thickness is calculated from the elastic or plastic theory ofbending.

Backing plates are used to strengthen flanges of columns as shown in Fig. 6.3, EN1993-1-8 (2005). In particular, they are used to strengthen T-stubs and methods aregiven in Table 6.2, EN 1993-1-8 (2005). The limits on dimensions of the plates aregiven in cl 6.2.4.3, EN 1993-1-8 (2005).

7.5.1 Plastic Methods for Plates (cl 6.2.4, EN 1993-1-8 (2005))

Parts of connections may be idealized as T-stubs as shown in Fig. 7.14 where theexternal force (Ft) is balanced by bolt force (Bt) and prying force (Q). A prying forceis an additional axial tensile force that is induced in a bolt due to the flexing andreaction of components.

There are three possible conditions of equilibrium for a T-stub at the ultimate limitstate (Fig. 7.15).

Case 1: Bolt failure as shown in Fig. 7.15(a)

Resolving forces vertically

Ft + �Bt = 0 (7.15)

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Ten

sile

forc

e in

the

bolt

(Bt)

Pre-loaded

Prying force Qat bolt failure

B t = F t

Prying force Qat the elastic stageof behaviour

Bolt withno preload

Applied external force (Ft)

(a) Relationship between external force and bolt force for a T-stub

(b)

Q QB t B t

F t

F t

m e

FIGURE 7.14 Prying forces for T-stubs

Bolt failure Bolt failure

Thick plates Elastic plates Plastic hinges

F t = ΣB t

Ft = ΣB t

Elastic boltPrying force

(a) No Prying force (b) Forms of prying force failure (c)

F t = Ft =

ΣB t /2 ΣBt /2 ΣBt /2 ΣBt /2

2Mpl + eΣB t

(m + e)

CQ Q Q

+ Q

B Q

A

4Mpl

m

Mpl Mpl

Mpl MplIdealized failure modes

Ft

2+ Q

Ft

2

FIGURE 7.15 Failure modes of T-stub flanges

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In this case the prying force (Q) is zero, the thickness of plate is a maximum and theforce in the bolt is a minimum (Table 6.2, EN 1993-1-8 (2005), mode 3).

Case 2: Bolt failure with partial yielding of the flange as shown in Fig. 7.15(b)

Taking moments of forces about C

−Mpl + Ft

2(m + e)+ �Bt

2e= 0

Rearranging, the design tension resistance

Ft = 2Mpl + �Btem + e

(7.16)

Prying force

Q = �Bt/2m − Mpl

m + e(7.17)

This is the case that is most likely to occur in practice because plate thicknesses arelimited to those available (Table 6.2, EN 1993-1-8 (2005), mode 2).

Case 3: Complete yielding of the flange as shown in Fig. 7.15(c)

Taking moments of forces about A

−Mpl + (Ft/2 + Q)m − Q(m + e) = 0 (i)

Taking moments about B

Mpl − Qe = 0 (ii)

Combining (i) and (ii) to eliminate Q

Ft = 4Mpl

m(7.18)

They prying force

Q = Mpl

e(7.19)

This case results in the smallest thickness of plate but the largest bolt force (Table 6.2,EN 1993-1-8 (2005), mode 1).

If needed the prying force can be calculated more accurately (Holmes and Martin,1983) and the theory has been shown to agree with the experimental results (Bahiaet al., 1981). Prying forces can be avoided by using non-flexible components or by theuse of stiffeners.

7.5.2 Plastic Method for the Thickness of Flange Plates

When considering T-stubs the length of the flange is known, but when analysing columnflanges in similar situations (Fig. 7.16(a)) an effective length needs to be determined.

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The method of determining the effective length is illustrated by a simple example forthe plastic yield lines shown in Fig. 7.16(b).

For a single bolt force then from virtual work

Ftm�

m + e=(

t2pfyp

4

)�

[4(m + e)

x+ 2x

m + e

](i)

Differentiating with respect to x to determine the value of x for which Ft is a minimum

∂Ft

∂x= −4

(m + e)x2 + 2

m + e= 0 hence x = (m + e)

√2 (ii)

Combining Eqs (i) and (ii) and rearranging, the thickness of the plate

tp =√(

Ftmfyp√2(m + e)

)(7.20)

From Eq. (i) the effective length

leff,b =[

4m + ex

+ 2xm + e

](m + e) = 4

√2(m + e)

M

N

m em e

m e

m e

x

pp

(a) (b) Simple theory (c) leff,b � 4 m � 1,25e(inner and end bolts)

(e) leff,b � 2 πm(inner and end bolts)

(d) leff,b � p (inner bolts) � 0,5p � 2 m � 0,625 e (outer bolts)

fΔ

FIGURE 7.16 Effective lengths for column flanges

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The work Eq. (i) can therefore be written as an equilibrium equation

Ftm =(

t2p fyp

4

)leff,b (7.21)

The above theory assumes simple idealized conditions and ignores, welds, washers,limited rotation at the plastic hinge and strain hardening. However the simple exampleshows the basic theoretical approach. Other solutions are related to research work(Stark and Bijlaard, 1988).

In practice the yield lines are more complicated (Figs 7.16(c), (d) and (e)) and moredifficult to analyse. Analysis using yield lines is avoided in the European Code bygiving the effective length of plate (leff,b) (Fig. 7.16) and applying the T-stub equations(Table 6.4, EN 1993-1-8 (2005)).

7.5.3 Elastic Methods for Plates

(a) Base plates are used to distribute the load from column sections as shown by thearea enclosed by the dotted line in Fig. 7.17. Elastic stress distribution with themaximum stress at the yield strength is used at the ultimate limit state to ensurethat large displacements do not occur. If the load is axial the pressure (fj) beneaththe base plate is uniform and the projection (c) of the steel plate beyond the edgeof the column, then for a cantilever from the simple theory of elastic bending atfirst yield per unit width

M = fyW

fjc2

2= fyt2

6

rearranging, the projection of the plate (Eq. (6.5), EN 1993-1-8 (2005))

c = t[

fy(3fjγM0)

]1/2

(7.22)

Justification for this theory is given in Holmes and Martin (1983).

c

fj per unit area

t

FIGURE 7.17 Bending strength of a column base plate

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Gusset plate

(a) Column base

Gusset plate

(b) Bracket

o

sg Fu

fgy

Edge stress depends onslenderness ratioApproximate buckling stress

distribution where Lg/ig � 185

osg Fu

fgy

Edge stress dependson slenderness ratioBuckling stress

distribution

Theoreticalhinge

(e)

(d)

o

Shape of deflected free edgeobserved in experiments for

Hg Lg

Hg � Lg

Hg

δbg

Bg

bg

Element strutTheoretical

hinge

sg

Lg

Fu

(c)

Theoretical model (Martin)

FIGURE 7.18 Gusset plates

(b) Gusset plates are used to stiffen base plates and brackets as shown in Fig. 7.18. Thefollowing theory for the buckling of a gusset plate is based on experimental work(Martin, 1979; Martin and Robinson, 1981). No advice is given in the EuropeanCode.

The basic structural unit is a triangular plate with loading applied to one edge asshown in Fig. 7.18(c). For theoretical purposes the plate is assumed to be composedof a series of fixed ended struts parallel to the free edge. The distribution of directstress across the width bg is shown on an element of the gusset plate in Fig. 7.18(d).The buckling stress varies depending on the slenderness ratio of the elemental strut.At the hinge the stress is for a very short strut (i.e. yield). At the free edge the value isfor the slenderness ratio of the elemental strut at the free edge.

For simplicity the buckling stress distribution shown in Fig. 7.18(d) can be replacedby a linear distribution as shown in Fig. 7.18(e), provided that the slenderness ratioof the free edge lg/ig < 185. This restraint is acceptable because slenderness ratios ofgusset plates in structural engineering do not often exceed this value.

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Taking moments of forces about the theoretical hinge at O (Fig. 7.18(e)) and ignoringthe moment of resistance of the base plate as justified (Martin and Robinson, 1981).

Fusg =∫ Bg

0(bgfgtg)δbg (7.23)

For each strip the buckling stress ( fg) is linearly related to the slenderness ratio (lg/ig).The effective length lg = bg when Lg = Hg, and from experiments (Martin, 1979) this isapproximately correct when Lg = Hg. The buckling stress for each strip can thereforebe expressed as

fg = fgy

[1 −

(bg

185ig

)]provided that lg/ig < 185 (7.24)

Combining Eqs (7.23) and (7.24), integrating, expressing the radius of gyration asig = tg/(2 × 31/2), and rearranging

tg = 2Fusg

fgyB2g

+ Bg

80(7.25)

where from the geometry of the plate

Bg = Lg

[(Lg/Hg)2 + 1]1/2 (7.26)

The slenderness ratio of the gusset plate may be defined as the slenderness ratio of astrip of unit width parallel to the free edge. From this definition and Eq. (7.26)

lgig

= (2 × 31/2)Bg

tg= 2 × 31/2(Lg/tg)

[(Lg/Hg)2 + 1]1/2 (7.27)

This theory is for non-slender gusset plates, that is, for lg/ig < 185. The theory forslender gusset plates is given elsewhere (Martin, 1979).

7.6 JOINTS SUBJECT TO SHEAR FORCES

Two simple connections subject to shear forces are shown in Fig. 7.19. The forces in themembers are assumed to be axial and to act through the centroidal axes of the members.This is correct in some situations, for example the bolted joint shown in Fig. 7.19(a).However it is not correct for the welded lap connection shown in Fig. 7.19(b) becausethe eccentricity of the force produces a moment which results in distortion at ultimateload. It is not correct for a roof truss joint as shown in Fig. 7.24 because althoughthe centroidal axes intersect and there are axial forces in the members there are alsosecondary moments.

A further assumption for simple joints is that the external forces are distributed evenlyto the bolts or welds. This is not correct for long bolted and welded joints and allowancemust be made for this.

Ch07-H5060.tex 12/7/2007 14: 23 page 225


F F F

b

e

F

F FF F

FF

l

x x xxxxx

x��

� � � �

Angle

FlatTie

(a) Bolted connection

Distortion at ultimate load

(b) Welded connection

FIGURE 7.19 Joints subject to shear forces

The overlap distance (l) is important for simple joints. For bolted joints the minimumof two bolts and the required end distances generally ensure that the lap is sufficient.However for welds (Fig. 7.19) the greater strength may indicate that the lap distancecan be small, but it must be appreciated that there must be room for stop and startlengths and that stress concentrations can occur. The minimum lap length is generallynot be less than four times the thickness of the thinner part joined where the weld iscontinuous. The minimum length of weld is 40 mm or six times the throat thickness.

For a joint with side welds only the lap is generally not less than the width of the memberand there should be end returns of twice the leg length of the weld (cl 4.3.2.1(4), EN1993-1-8 (2005)) to reduce stress concentrations.

7.7 JOINTS SUBJECT TO ECCENTRIC SHEAR FORCES

Joints, such as shown in Fig. 7.20(a), are subject to eccentric shear forces which tendto rotate the joint. This produces a resultant shear force on a fastener (bolt, or unitlength of weld) from the direct shear force and the moment.

The forces acting on a group of fasteners can be idealized as shown in Fig. 7.20(b).The bolt group rotates about the theoretical instantaneous centre of rotation whichvaries in position depending on the magnitudes of the external forces V and H andthe eccentricity e. In the linear elastic stage of behaviour it is reasonable to assumethat the force acting on a fastener is proportional to the distance from the centre ofrotation. At ultimate load this assumption is not strictly correct but the error involvedis not great. For a rigorous solution to the theory and accuracy at ultimate load seeBahia and Martin (1980).

Ch07-H5060.tex 12/7/2007 14: 23 page 226


Vertical componentof force fromtorsional moment

Horizontal componentof force fromtorsional moment

V

V

H

H

G

M �Ve n

θ

θ z G

yG

rG

(c) Vector digram for simple vector addition theory for a typical fastener

Vn

o

y y

yn

y e

rn

r1

r2θ2

θn θ1

Zn

z

z

rn

r2 rn

r1

Fmax

Fmax

Fmax Fastener

Centroid offastener group

NH

(b) Diagram for general theory

(a)

FIGURE 7.20 Joints subject to eccentric shear forces

Although this is the correct approach to the theory there is a simpler more practicalmethod in common use which gives the same values. It is assumed that rotation occursabout the centroid of the fastener group and for convenience the forces acting on afastener are parallel to the z–z and y–y axes as shown in Fig. 7.20(c) There forces arecombined vectorially and the resultant force on a fastener furthest from the centre ofrotation is

FR = [F2y + F2

z ]1/2 =[(

Vn

+ MyG

Ix

)2

+(

Hn

+ MzG

Ix

)2]1/2

(7.28)

where

n is the number of fasteners in the group

yG and zG are coordinates of a fastener related to the centroid of the fastener group

IX = IY + IZ

IX , IY and IZ are second moments of area of unit size fasteners about the X–X, Y–Yand Z–Z axes

The method is used in practice for ordinary bolts, high strength friction grip bolts andwelds.

7.8 JOINTS WITH END BEARING

Some joints involve end bearing between components (Fig. 7.21). End bearing canoccur in beam-to-column (Fig. 6.15, EN 1993-1-8 (2005)), bracket-to-column, beam-to-beam and column-to-base joints.

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H

V

O

R

Ro

μRo

d p

d r M

Centre of rotation atstiff bearing O

(b)

(d)(c)

(a)

End bearing

End bearing

End bearing

FIGURE 7.21 Joints with end bearing

Where end bearing occurs, rotation takes place about a stiff axis of rotation, axis O–Oshown in Fig. 7.21(d). The reaction force Ro is generally large and the bearing mayhave to be reinforced if it is not to distort under the load. The balancing tensile forceR is resisted by bolts or welds. If there is slip at the stiff bearing a frictional force μRo

develops parallel to the stiff bearing surface.

Consider an end bearing joint subject to external forces V , H and moment M as shownin Fig. 7.21(d). If slip occurs resolving forces vertically

V − μRo < 0 (7.29)

Taking moments of forces about force R

M + −H(dr − dp) − Rodr = 0 (7.30)

where R is the resultant force of the fasteners acting at a distance dr from the axis ofrotation O–O.

Combining Eqs (7.29) and (7.30) to eliminate Ro, then slip will not occur if

μ[M ± H(dr − dp)]Vdr

> 1 (7.31)

Most joints with end bearing do not slip and therefore the fasteners are not subjectto the external shear force. One exception is a bracket supporting a load with a smalleccentricity.

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In the elastic stage of behaviour it is assumed that the forces acting on a fastener areproportional to the distance from the axis of rotation O–O. If, conservatively, this isalso assumed to occur at ultimate load then:

Taking moment of forces about axis O–O the maximum tensile force resisted by afastener

Ft(max) = [M ± Hdp]zmax

IO(7.32)

where

IO = �z2 is the second moment of area of unit size fasteners about axis O–O Resolvingforces vertically the shear force on a fastener

Fs = (V − μRo)n

={

V − μ[

Mdr

± H(1 − dp/dr)]}

n(7.33)

This equation is formed assuming that bolts of the same size and design strength resistequal shear forces and also that slip has taken place.

If the fasteners are welds then Ft(max) and Fs are combined vectorially. If the fasten-ers are ordinary bolts the forces are combined using Eq. (7.8) and related to designstrengths. If friction grip bolts are used the forces are combined using Eqs (7.13) and(7.14) and related to design strengths.

Traditional design methods ignore the existence of the frictional force which errs onthe side of safety. However, research (Bahia et al., 1981) has shown that the frictionalforce does resist part of the shear force. If pre-loaded bolts are used then slip mayoccur at service load or at ultimate load.

7.9 ‘PINNED’ JOINTS (CL 3.13, EN 1993-1-8 (2005))Some simple joints (e.g. a tie bar) are connected by real pins as shown in Fig. 7.22(a).Provided that the pins are not corroded, or blocked with debris, they will act as pinjoints, that is, they will resist forces but not moments. Tie bars are rarely used nowbecause of the cost of manufacture, risk of seizure from corrosion or debris, andbecause safety depends on a single pin.

Other connections shown in Fig. 7.22 are designated as ‘pins’ because the rotationalrestraint is small. In the past these joints have been designed assuming that the rota-tional resistance is zero and the connection resists direct forces only. The generalapproach to design of these ‘pinned’ connections follows.

7.9.1 Pinned Beam-to-Column Joints (Figs 7.22(b), (c),(e) and (f))

In design calculations for the joint shown in Fig. 7.22 (b) it is assumed that the shearforce is resisted by the four bolts connecting the bottom cleat to the column flange.

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Shear force

Tie force

Plate welded to end of beam

(e) Beam-to-column

(c) Beam-to-column

(b) Beam-to-column

(a) Tie rod

(f) Beam-to-column

(d) Beam-to-beam

(h) Column-to-foundation(g) Beam-to-beam

Shear force

Tie force

Plate welded to column

Shear force

Tie force

Plate welded to beam

Weld

Shear force

Tie force

Angle section

Shear force

Tie force

Angle section

Angle section

Shear force

Tie force

Angle section

Main beam

Secondary beam

Axial force

Pin joint

Bolted ‘pin’ connections

Welded-bolted ‘pin’ connections

Axial force

Base plate welded to column

Reinforced concretefoundation

FIGURE 7.22 Examples of ‘pinned’ joints

During erection the bottom cleat, which is bolted or welded to the column, is used asa marker by the crane operator when the beam is placed. The top cleat is assumed toresist no vertical load but it does provide torsional resistance which is important forlateral stability. The top and bottom cleats also resist the tie force. The resistance of

Ch07-H5060.tex 12/7/2007 14: 23 page 230


the web of the beam to shear, bearing and buckling must be checked. At ultimate loadthe rotation at the end of the beam often introduces end moments. In practice, theseare assumed to be small and are ignored.

Other types of ‘pinned’ beam-to-column joints are shown in Figs. 7.22(c), (e) and (f)where the depth of the joint is kept to a minimum. For example, the end plate depth forFig. 7.22(e) is kept to a minimum to reduce end moments and the empirical thicknessis 8 mm for UB sizes up to 457 × 191 kg, and 10 mm for sizes greater than 533 × 210 kg.Examples of ‘pin’ joints used in practice are given by Pillinger (1988).

Where the depth of the connection for Fig. 7.22(e) is greater and is also slip resistantthen advice on the distribution of forces to the bolts is given in Fig. 6.15, EN 1993-1-8(2005).

7.9.2 ‘Pinned’ Beam-to-Beam Joints (Figs 7.22 (d) and (g))

The transverse secondary beam is connected to the main beam through angle cleatsas shown in Fig. 7.22(d). It is assumed in design that the shear force is transferred tothe main beam via the bolts in the web of the main beam. These bolts are thereforedesigned for single shear and bearing on the web of the main beam and on the anglecleats. It follows therefore that the shear force is eccentric to the bolts in the webof the secondary beam. These bolts are double shear and bearing on the web of thesecondary beam and the angle cleats.

Steel fabricators and erectors often prefer a welded and plate (Fig. 7.22 (g)) as analternative to the angle cleats shown in Fig. 7.22 (d). This results in a more rigid jointand an end moment is introduced to the end of the secondary beam which is dependanton the torsional stiffness of the main beam. If there are secondary beams on both sidesof the main beam the secondary moment can be large.

7.9.3 ‘Pinned’ Column-to-Foundation Joints (Fig. 7.22 (h))

The column is fastened to the base plate which is connected to the foundation byfoundation bolts. This type of joint is used where the predominant force in the col-umn is axial but there is generally a small shear force. The size of the foundationbolt is based on resisting the forces, but with a minimum size of M16. The thicknessof the base plate is related to pressure beneath the base and the cantilever effectthe plate.

7.10 ‘RIGID’ JOINTS

‘Rigid’ (or ‘fixed’) joints (Fig. 7.23) exhibit small rotational displacements in the elasticstage of behaviour. They are now more common as connections to members are weldedin the workshops and bolted on site. They are useful to limit deflections of members,

Ch07-H5060.tex 12/7/2007 14: 23 page 231


Strap

RHS sections

Haunch

End plateWeb stiffener

M

Ro

d

(j)(i)

(h)

(f)(e)(d)

(a) (b) (c)

(g)

FIGURE 7.23 Examples of ‘rigid’ joints

resist fatigue and resist impact loading. However, generally, the design procedure forthe frame requires global analysis and the components are more highly stressed.

7.10.1 ‘Rigid’ Column Bracket Joints (Figs 7.23 (a), (b) and (c))

For the brackets shown in Figs 7.23 (a) and (c) no end bearing is involved whereas thebracket shown in Fig. 7.23 (b) involves end bearing.

Ch07-H5060.tex 12/7/2007 14: 23 page 232


7.10.2 ‘Rigid’ Beam-to-Column Joints (Figs 7.23 (d) and (e))

The most rigid type of joint is where the beam is welded directly to the column onsite, as shown in Fig. 7.23(d), but this is expensive and it is difficult to control thequality of the weld. Alternatively stub cantilever beams can be welded to the columnin the workshops and a suspended beam site bolted between the ends of the can-tilever. The bolted connection is positioned as close to the point of contraflexure aspossible.

The method commonly used in Britain is to weld end plates to the beam in the work-shops and to bolt these to the columns on site as shown in Fig. 7.23(e). The numbersof bolts is usually six, as shown, because of the limited depth available. If the momentof resistance needs to be increased then it is necessary to increase the lever arm byhaunching the beam at the end. This connection is less rigid than welding the beamdirectly to the column but it is easier to manufacture and erect. The amount of rota-tion depends on the thickness of the end plate, thickness of the column flanges andthe extensibility of the bolts. Advice on details is given in cl 6.2.7.2, EN 1993-1-8(2005).

If the connection is close to a plastic hinge then it must be decided whether the plastichinge should forming the beam, or the column, or the connection. Recent researchfavours the formation of the hinge in the beam and therefore the column and theconnection must be overdesigned.

End bearing occurs between beam and column and the first step in design is to checkwhether slip occurs using Eq. (7.31). Generally, because the bending moment is large,slip does not occur and the size of the top four bolts required can be determinedapproximately by taking moments of forces about the compression flange of the beam.The tensile force in a bolt

Ft = M4(h − tf )

(7.34)

The tensile force in a bolt is increased by a prying force. Equation (7.34) assumes thatthe two bolts close to the compression flange of the beam do not resist any part ofthe bending moment. Lever arm recommendations are given in Fig. 6.15 (EN 1993-1-8 (2005)), The thicknesses of the end plate is determined by assuming equivalentT-stubs and associated yield lines. This method can be compared with a survey ofexisting literature on end plates by Mann and Morris (1979) who recommended

⎡⎢⎣ M

dbf fy( 4wp

sv+ dbf

sh

)⎤⎥⎦

1/2

< tp <

[Mbp

2wpdbf fy

]1/2

(7.35)

provided that Bp = 9db, sh = 6db, sv = 6db and eb > 2,5db.

Ch07-H5060.tex 12/7/2007 14: 23 page 233


The thickness of the column flange can also be determined from equivalent T-stubswith yield line patterns. For comparison a further survey by Mann and Morris (1979)recommended:

For unstiffened column flanges the thickness of the flange

0,28[

Mdbf fy

]1/2

< tcf < 0,39[

Mdbf fy

]1/2

(7.36)

For stiffened column flanges the thickness of the flange

0,23[

Mdbf fy

]1/2

< tcf < 0,32[

Mdbf fy

]1/2

(7.37)

Equations (7.36) and (7.37) are valid provided that bc = 2,5db and cc = tst + 5db. Wherethe column flange thickness is inadequate backing plates can be used.

The force Ro at the axis of rotation may produce failure by bearing or buckling of theweb of the column. If failure is likely to occur then stiffeners can be welded into theweb of the column. However, stiffeners increase costs and reduce the room availablefor bolts. The tensile force balancing Ro (Fig. 7.21) which acts on the web of the columnis often not critical but the strength of the column web must be checked.

The size of the fillet weld connecting the end plate to the beam is determined by assum-ing rotation about the axis O–O at the bottom flange of the beam, but an alternativemore conservative method is to assume rotation about the centroidal axis of the weld.

7.10.3 ‘Rigid’ Beam Splices (Fig. 7.23(g))

Beam splices are introduced to extend standard bar lengths, or to facilitate constructionand transport. Splices are generally located at sections where the forces are minimum,and to avoid local geometrical deformations of the structure pre-loaded bolts are used.The connection is usually made on site and therefore bolts are used.

A simple design method is to assume that the entire shear force is resisted by the websplice and the flanges resist the entire bending moment. These assumptions are notcorrect but it simplifies the design and the errors are generally not large. Alternativelypart of the moment is assumed to be resisted by the web. The bolts in the web, indouble shear, are subject to an eccentric shear load. The bolts in the flange are also indouble shear for two flange plates and the force on a bolt is obtained a simple momentequation F = M/(nd).

7.10.4 ‘Rigid’ Column-to-Column Joints (Fig. 7.23(f))

Column splices are used to extend standard bar lengths, to facilitate erection andtransport, and for economy by reducing the section size. Where bending moments arelarge then, as with beam splices, pre-loaded bolts are used to maintain the axial line of

Ch07-H5060.tex 12/7/2007 14: 23 page 234


the column. Where sections change size steel packings and an end plate are requiredto ensure that a good fit is obtained.

Generally the ends of the columns are in contact, or in contact with the end plate, andtherefore end bearing occurs. Any shear force will be resisted by the friction at the endbearing and the web plates (or cleats). Where forces are small sizes of componentsare decided from experience, practicality and corrosion resistance.

7.10.5 ‘Rigid’ Hollow Section Joints (Fig. 7.23(h))

This is a typical ‘T’ joint for a Vierendeel girder where members of different widthintersect. If forces and moments are not too large then the connection can be madewithout using stiffeners. Rectangular hollow sections are also used in braced triangu-lated trusses but it is difficult to arrange for the centre lines of the members to intersectat a point. The offsets introduce moments which should be taken into account in theanalysis of the structure (Purkiss and Croxton, 1981). Further information on analysisand design of connections is given by Davies (1981).

Circular hollow sections are also used for trusses but the geometry of the connectionis more difficult and connections are more expensive to form. A review of methodsof analysis of the strength of these connections is given by Stamenkovic and Sparrow(1981).

The failure modes for hollow section joints are (cl 7.2.2, EN 1993-1-8 (2005)):

(a) chord face failure,(b) chord side wall failure,(c) chord shear failure,(d) punching shear failure,(e) brace failure,(f) local buckling.

The design resistance of numerous types of joints are shown in Tables 7.2–7.24, EN1993-1-8 (2005). These take into account in plane and out of plane forces.

Guidance on strength of welds for connecting members is given in cl 7.5.1, EN1993-1-8 (2005). These are only applicable if they meet the joint parameters of Table7.8, EN 1993-1-8 (2005).

7.10.6 ‘Rigid’ Column-to-Foundation Joints (Fig. 7.23(i))

Where bending moments are small a simple slab base is satisfactory with the bolts inline with the column axis. As the bending moment increases the bolts are off-set fromthe column axis. A built-up base is used where moments are large. The basic designmethod is based on ‘T’ stub theory (cl 6.2.8, EN 1993-1-8 (2005)).

Ch07-H5060.tex 12/7/2007 14: 23 page 235


For a gussetted slab base subject to an axial load and bending moment it is assumedthat at ultimate load the distribution of stresses beneath the steel base plate are asshown in Fig. 7.34(c). The depth of the compression zone is determined approximatelyby taking moments of forces about the tensile bolts

x =[

Md + N

2

]bc fij

(7.38)

where bc is based on the cantilever length (c) and fij is the allowable bearing pressure.

Alternatively it can be assumed that the centroid of the compression zone is locatedunder the column flange (Table 6.7, EN 1993-1-8 (2005)).

From taking moments of forces about the compressive force the tensile force in a bolt is

Ft =[

Md − N

2

]n

(7.39)

The thickness of the base plate is found based on elastic bending of the cantileverlength (c).

For the built-up base the base plate thickness is determined by the same method. Thegusset plate size is determined by the method shown in Section 7.5.3. The size of thewelds connecting the gusset plate to the base and to the column can be determinedassuming end bearing. Alternatively end bearing can be ignored and a larger sizedetermined by assuming rotation about the centroid of the weld group.

7.10.7 ‘Rigid’ Knee Joint for a Portal Frame (Fig. 7.23(j))

This type of joint is similar to the beam-to-column connection shown in Fig. 7.23(e).However because portal frames are often designed using plastic analysis the magnitudeof the reaction Ro = M/d is generally high and consequently the shear stresses in theweb are close to the limit. The magnitude of Ro is reduced by haunching the beamand consequently increasing the distance d, but web stiffeners are generally requiredfor the column. To reduce the shear deformation in the column web stiffeners can beintroduced as described and tested by Morris and Newsome (1981).

The tensile force which balances Ro can be resisted by a strap, or by a group of boltsthrough the flange of the column. The strap may interfere with the placing of purlins inthe roof and the alternative group of bolts may cause distortion of the column flangesif the column flange thickness is small. The distortion can be controlled by the use offlange stiffeners, or increase the size of the column section, but both increase the cost.

EXAMPLE 7.2 Design of a ‘pin’ joint for a roof truss (Fig. 7.24). The forces, sizeof angles and tees have been obtained from an analysis at ultimate load assuming pinjoints and axial forces in the members (cl 5.1.5(2), EN 1993-1-8 (2005)).

Ch07-H5060.tex 12/7/2007 14: 23 page 236


310

125 kN215 kN

310

30kN

30 kN100 kN

60 � 60 � 6 angles125 � 75 � 8 angles

165 � 152 � 20 kg tee165 � 152 � 20 kg tee

28

31

1615

24

Gusset plate tg � 10

bg

347 33

0

151,

9

10,2

10

t � 6,1

60 80 40Strap6 mm fillet weldsGrade S275 steelM20 class 4.6 bolts

FIGURE 7.24 ‘Pinned’ roof truss joint

The centroidal axes of the members intersect so there is no eccentricity to be taken intoaccount (cls 2.7 and 3.10.3, EN 1993-1-8 (2005)). The thickness of the gusset plate is atleast 6 mm to resist corrosion, and at least equal to the minimum thickness of the angleor tee (6,1 mm). Use a 10 mm thick plate grade S275 steel. A rectangular plate is simpleto mark and cut, and low in fabrication cost. Alternatively a more complicated shapecan be used which is aesthetically more acceptable but the fabrication cost is greater.

Member 24, structural tee cut from a UB (165 × 152 × 20 kg) welded to a gusset plate,design force NEd = 215 kN.

Tensile resistance of the tee Grade S275 steel (Eqs (6.6) and (6.7), EN 1993-1-1 (2005))

Npl,Rd = AfyγM0

= 2580 × 2751,00 × 1E3

= 709,5 > 215 kN satisfactory.

Nu,Rd = 0,9Anetfu

γM2= 0,9 × (2580 − 2 × 22) × 430

(1,25 × 1E3)


Assuming a 6 mm fillet weld for Grade S275 steel (Eq. (4.4), EN 1993-1-8 (2005))

Fw,Rd = fu a(31/2βwγM2)

= 430 × 0,7 × 61E3

(31/2 × 0,85 × 1,25)= 0,981 kN/mm

Ch07-H5060.tex 12/7/2007 14: 23 page 237


Effective length of weld required to resist the tensile force using the simplified method(cl 4.5.3.3, EN 1993-1-8 (2005))

NEd

Fw,Rd= 215

0,981= 219,2 mm

Two side fillet welds of 110 mm length would be satisfactory but in practice the lengthswould probably be the full overlap (i.e. 2 × 310 = 620 mm).

Member 31, structural tee cut from a UB (165 × 152 × 20 kg) bolted to the gusset plateand strap, design force NEd = 125 kN.

Resistance of 4-M20 class 4.6 bolts in single shear (Table 3.4, EN 1993-1-8 (2005))

�Fv,Rd = nb

(0,6As

fub

γM2

)

= 4 ×[

0,6 × 245 × 4001,25 × 1E3

]= 188 > (NEd = 125) kN satisfactory.

Resistance of 2-M20 class 4.6 bolts in bearing on the web of the tee section (t = 6,1 mm)(Table 3.4, EN 1993-1-8 (2005))

�Fb,Rd = nb

(k1αbfubdt

γM2

)

= 2 ×[

2,5 × 0,909 × 400 × 20 × 6,11,25 × 1E3

]= 177,4 > (NEd = 125) kN satisfactory

where

for an end bolt αb = e1/(3do) = 60/(3 × 22) = 0,909

for an edge bolt k1 = 2,8e2/do − 1,7 = 2,8 × 55/22 − 1,7 = 5,3 > 2, 5

The strap increases the out of plane stiffness of the truss. Connections for othermembers can be designed by the same method.

EXAMPLE 7.3 ‘Rigid’ column bracket. Determine the size of the componentsrequired to connect the bracket to the column shown in Fig. 7.25 using Grade S355steel. The forces shown are applied to one gusset plate at ultimate load.

For the 10 bolts (Fig. 7.25(a)) of unit cross-sectional area the properties of the boltgroup are:

Second moment of area of the bolt group about the centroidal y–y axis

Iy = �(∂A)z2 = 4(802 + 1602) = 128E3 mm4

Second moment of area of the bolt group about the centroidal z–z axis

Iz = �(∂A)y2 = 10(70)2 = 49E3 mm4

Ch07-H5060.tex 12/7/2007 14: 23 page 238


bf � 255,9Lg � 282,95

tg � 104 at

80

� 3

20

Hg

� 4

00

tf � 17,3

Bg � 231

ev

� 3

50

V � 210 kN

H � 45 kN

eh � 250

140

y y

254 � 254 � 89 kg UC

z

z

G

sg

4040

Gussetplate

Grade S 355 steel

Lg � 225bw � 200

tg � 10

V � 210 kN

H � 45 kN

h � 228,6

eh � 250

e v �

350

Hg

� d

w �

400

sg � 150

h g �

150

yy

2 � 229 � 76 channels

z

z

G

Gussetplate

Bg � 196,1

(a) Bracket bolted to a UC (b) Bracket welded to a compound column

FIGURE 7.25 ‘Rigid’ column brackets

Second moment of area of the bolt group about the centroidal polar x–x axis

Ix = Iy + Iz = (128 + 49)E3 = 177E3 mm4

From Eq. (7.28) the maximum vector force in the direction of the z–z axis on a boltfurthest from the centroid of the bolt group

Fz = Vnb

+ (Veh + Hev)yn

Ix

= 21010

+ (210 × 250 + 45 × 350)70177E3

= 48 kN

The maximum vector force in the direction of the y–y axis on the same bolt

Fy = Hnb

+ (Veh + Hev)zn

Ix

= 4510

+ (210 × 250 + 45 × 350)160177E3

= 66,2 kN

Resultant vector design force on this bolt

Fr = (F2z + F2

y )1/2 = (482 + 66, 22)1/2 = 81, 77 kN

Solution (a) using class 4.6 bolts. Shear resistance of a M30 class 4.6 bolt in single shear(Table 3.4, EN 1993-1-8 (2005))

Fv,Rd = 0,6Asfub

γM2

= 0,6 × 561 × 4001,25 × 1E3

= 107,7 > (Fr,Ed = 81, 77) kN satisfactory.

Ch07-H5060.tex 12/7/2007 14: 23 page 239


However the recommended maximum bolt diameter for a column flange width of254 mm is 24 mm (Annex A4). Use a higher class of bolt.

Solution (b) using M20 class 8.8 bolts not pre-loaded. Shear resistance of an M20 class8.8 bolt in single shear (Table 3.4, EN 1993-1-8 (2005))

Fv,Rd = 0, 6Asfub

γM2

= 0,6 × 245 × 8001,25 × 1E3

= 94,1 > (Fr,Ed = 81,77) kN satisfactory.

M20 class 8.8 bolt in bearing on the gusset plate (t = 10 mm) (Table 3.4, EN 1993-1-8(2005)).

Bearing strength

Fb,Rd = k1αbfupdtγM2

= 2,5 × 0,606 × 510 × 20 × 101,25 × 1E3

= 123,6 > (Fr,Ed = 81, 77) kN

where


for an edge bolt k1 = 2,8e2/do − 1,7 = 2,8 × 58/22 − 1,7 = 5,68 > 2,5

Solution (c) using pre-loaded M22 class 10.9 bolts (cl 3.9, Eqs (3.6) and (3.7), EN1993-1-8 (2005))

Fs,Rd = ksnμbFp,C

γM3

= 1,0 × 1,0 × 0,5 × 0,7 × 1E3 × 3031E3

1,25= 84,8 > (Fr,Ed = 81,77) kN

To determine the thickness of the gusset plate for the bolted joint Fig. 7.25(a)

Lg = 225 + (255,9 − 140)2

= 282,95 mm

sg = 150 + (255,9 − 140)2

= 207,95 mm

Width of the gusset plate perpendicular to the free edge (Eq. (7.26))

Bg = Lg[( Lg

Hg

)2 + 1]1/2 = 282,95[(

282,95400

)2 + 1]1/2 = 231,0 mm

Ch07-H5060.tex 12/7/2007 14: 23 page 240


From Eq. (7.25), replacing the term (Pusg) with (Vsg + Hhg), the thickness of the gussetplate Grade S355 steel

tg = 2(VSg + Hhg)(fgyB2

g

γM1

) + Bg

80

= 2 × (210 × 207,95 + 45 × 150)E3(355×2312

1,0

) + 23180

= 8, 21 mm; use a 10 mm thick plate of Grade S355 steel

Check the slenderness ratio of the gusset plate (Eq. (7.27)).

lgig

= 2 × 31/2 Bg

tg= 2 × 31/2 × 231

10

= 80,02 < 185 the limit of the slenderness ratio for the application of the

theory, satisfactory.

Solution (d) using welds (Fig. 7.25(b)).

Where it is not possible to bolt to a column, for example the compound channel columnshown in Fig. 7.25(b), then welds are used. The connection is rigid and for welds ofunit size the properties of the weld group are:

Total length of weld

Lw = 2(dw + bw) = 2(400 + 200) = 1200 mm

Second moment of area of the weld group about the centroidal y–y axis

Iy = � (∂A) z2 = 2

[dw3

12+ bw

(dw

2

)2]

= 2

[4003

12+ 200

(4002

)2]

= 26, 67E6 mm4

Second moment of area of the weld group about the centroidal z–z axis

Iz = �(∂A)y2 = 2

[b3

w12

+ dw

(bw

2

)2]

= 2

[2003

12+ 400

(2002

)2]

= 9, 33E6 mm4

Second moment of area of the weld group about the centroidal polar x–x axis

Ix = Iy + Iz = (26, 67 + 9, 33)E6 = 36E6 mm4

Ch07-H5060.tex 12/7/2007 14: 23 page 241


Maximum vector force in the direction of the z–z axis on a weld element furthest fromthe centroid of the weld group from Eq. (7.28).

Fz = VLw

+ (Veh + Hev)yn

Ix

= 2101200

+ (210 × 250 + 45 × 350)10036E6

= 0,365 kN/mm

Maximum vector force in the direction of the y–y axis on the same weld element

Fy = HLw

+ (Veh + Hev)zn

Ix

= 451200

+ (210 × 250 + 45 × 350)20036E6

= 0,417 kN/mm

Resultant vector design force on this weld element

Fr,Ed = (F2z + F2

y )1/2 = (0, 3652 + 0, 4172)1/2 = 0, 554 kN/mm


Fw,Rd = fua

(312 βwγM2)

= 430 × 0,7 × 61E3

(31/2 × 0,85 × 1,25)

= 0,981 < (Fr,Ed = 0, 554) kN/mm satisfactory.

To determine the thickness of the gusset plate for the welded joint Fig. 7.25(b).

From Eq. (7.26) the width of the gusset plate perpendicular to the free edge

Bg = Lg[( Lg

Hg

)2 + 1]1/2 = 225[(

225400

)2 + 1]1/2 = 196, 1 mm

From Eq. (7.25) replacing the term Pusg by (Vsg + Hhg)

tg = 2(Vsg + Hhg)(fgyB2

g

γM1

) + Bg

80

= 2 × (210 × 150 + 45 × 150)E3(3551,0 × 196,12

) + 196, 180

= 8, 05 mm, use an 10 mm thick plate of Grade S355 steel.

Check the slenderness ratio of the gusset plate from Eq. (7.27)

lgig

= 2 × 31/2Bg

tg= 2 × 31/2 × 196, 1

10

= 67, 93 < 185 the limit of the slenderness ratio for the application of this theory.

Ch07-H5060.tex 12/7/2007 14: 23 page 242


End shear force onbeam V � 225 kN

Top angle cleat 80 � 80 �10 mm anglesupports no vertical load

203 � 203 � 86 UC

457 � 191 � 98 kg UB

Tie force

75 kN

Clearance 5 mm

tf � 20,5 tf � 19,6tw � 13

h �

467

,4

tw � 11,4r � 10,2

bf � 208,8

b � 192,8

Grade S 355 steel

M20 class 8,8 bolts

Bottom cleat 125 � 75 � 10 mm anglesupport all vertical load

FIGURE 7.26 ‘Pinned’beam-to-column joint

EXAMPLE 7.4 ‘Pinned’ beam-to-column connection. Check the size of componentsfor the connection shown in Fig. 7.26 at ultimate load.

(a) If the design shear force of Fv,Ed = 225 kN is resisted by 4-M20 grade 8.8 bolts inthe bottom cleat (125 × 75 × 10 mm angle), then the design shear force per bolt

Fv,Ed = 2254

= 56,25 kN

Shear resistance of an M20 grade 8.8 bolt (Table 3.4, EN 1993-1-8 (2005))

Fv,Rd = 0,6Asfub

γM2

= 0,6 × 245 × 8001,25 × 1E3

= 94,1 > (Fv,Ed = 56,25) kN satisfactory.

Bearing resistance of an M20 class 8.8 bolt bearing on the leg of the angle(ta = 10 mm) (Table 3.4, EN 1993-1-8 (2005)).

Fb,Rd = k1αbfuadtaγM2

= 2,12 × 0,682 × 510 × 20 × 101,25 × 1E3

= 118 > (Fv,Ed = 56,25) kN satisfactory.

where for an end bolt αb = e1/(3do) = 45/(3 × 22) = 0,682and for an edge bolt k1 = 2,8 (e2/do) − 1,7 = 2,8 × (30/22) − 1,7 = 2,12 < 2,5

Assume the 4-M20 class 8.8 bolts connecting the bottom cleat to the columnwhich resist the shear force of 225 kN also resist the tensile force of 75 kN. Designtensile force per bolt from the 75 kN tie force is

Ft,Ed = 754

= 18,75 kN

Tensile resistance of an M-20 class 8.8 bolt

Ft,Rd = 0,9Asfu

γM2= 0,9 × 245 × 800

1,25 × 1E3= 141,1 kN

Ch07-H5060.tex 12/7/2007 14: 23 page 243


Combined shear and tension for a bolt (Table 3.4, EN 1993-1-8 (2005))

Fv,Ed

Fv,Rd+ Ft,Ed

(1,4Ft,Rd)

= 56,2594,1

+ 18,75(1,4 × 141,1)

= 0,693 < 1 satisfactory.

(b) Alternatively the four bolts in the vertical leg of the bottom angle could be replacedby fillet welds along the two vertical edges of the angle.Resistance of two 6 mm fillet welds for Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005))

2lwFw,Rd = 2lwfua(31/2βwγM2)

= 2 × 125 × 510 × 0,7 × 61E3

(31/2 × 0,9 × 1,25)= 275 > (VEd = 225) kN.

The top cleat is used to provide torsional resistance against lateral buckling of thebeam and to resist tie and erection forces. The angle must be at least 6 mm thickto resist corrosion and the leg of sufficient length to accommodate M20 bolts. FromSection Tables a 80 × 80 × 10 mm angle is chosen.

For resistance to transverse shear forces for the beam see Example 4.12

EXAMPLE 7.5 ‘Pinned’ beam-to-beam connection. Determine the size of the com-ponents required for the connection shown in Fig. 7.27. The beam sizes have beendetermined from bending calculations at ultimate load.

Assuming that the M20 class 4.6 bolts through the web of the main beam B (GradeS275 steel) are subject to single shear forces.

Shear resistance of a M20 class 4.6 bolt in single shear (Table 3.4, EN 1993-1-8 (2005))

Fv,Rd = 0,6Asfub

γM2= 0,6 × 245 × 400

1,25 × 1E3= 47,0 kN

M20 class 4.6 bolt in bearing on the web of the transverse beam A (tw = 6,9 mm)(Table 3.4 EN 1993-1-8 (2005)).

For an end bolt αb = e1/(3do) = 40/(3 × 22) = 0,606

For an edge bolt k1 = 2,8e2/do − 1,7 = 2,8 × 30/22 − 1,7 = 2,12 < 2,5

Fb,Rd = k1αbfubdtwγM2

= 2,12 × 0,606 × 400 × 20 × 6,91,25 × 1E3

= 56,7 > (Fv,Rd = 47,0) kN.

nb = VEd

Fv,Rd= 150

47,0= 3,19 use 4-M20 class 4.6 bolts.

Ch07-H5060.tex 12/7/2007 14: 23 page 244


Transverse beam A356 � 171 � 45 kg UB

Main beam B610 � 305 � 149 kg UB

End shear forceV �150 kN

bf � 171,0

bf � 304,8

t f �

19,

7

tw � 6,9

tw � 6,9

tw � 11,9 t f �

9,7

h �

352

,0

h �

609

,6

70 � 70 � 10 angles

Grade S275 steelM20 class 4,6 bolts

End elevationSide elevation

158

3840

8 mm

40

40

z

z

y y

b

6565

160

a

3 �

70

�21

0

Clearance

FIGURE 7.27 ‘Pinned’ beam-to-beam joint

Assuming that the bolts connecting the angle cleats to the web of the transverse beamA are in double shear and subject to an eccentric load.

Second moments of area of the bolt group about the centroidal axis for bolts of unitarea are

Iy = �(∂A)z2 = 2(352 + 1052) = 24,5E3 mm4

Iz = 0

Ix = Iy + Iz = 24,5E3 mm4

Maximum shear force on a bolt in the y direction from Eq. (7.28)

Fy = VEdezmax

Ix= 150 × 40 × 105

24,5E3= 25,71 kN

Average shear force on a bolt in the z direction

Fz = VEd

nb= 150

4= 37,5 kN

Maximum resultant design shear force on a bolt

Fr,Ed = (F2y + F2

z )1/2 = (25,712 + 37,52)1/2 = 45,47 kN

Double shear strength of an M20 class 4.6 bolt

2Fv,Rd = 2 × 47,0 = 94,0 > (Fr,Ed = 45,47) kN, satisfactory.

Ch07-H5060.tex 12/7/2007 14: 23 page 245


Check ‘block shear tearing’ (cl 3.10.2, EN 1993-1-8 (2005)) for line of holes in end oftransverse beam A

Veff,2,Rd = 0,5fuAnt

γM2+ 1

31/2 fyAnv

γMO

= 0 + 131/2 × 275 × 1194

1,0 × 1E3= 189,6 > (VEd = 150) kN satisfactory.

which includes

Anv = (3 × hole spacing + end distance − 3,5 × hole diameter)tw

= (3 × 70 + 40 − 3,5 × 22) × 6,9 = 1194 mm2

EXAMPLE 7.6 ‘Pinned’ column-to-foundation connection. Determine the size ofthe components for the axially loaded base shown in Fig. 7.28 at the ultimate limitstate. Concrete cylinder crushing strength fck = 20 MPa.

Lp � 450

tf � 20,5

t w �

13

b f �

258

,3

Bp

� 4

2,5

h � 266,7 C

C

C

C 80

120

N = 1500 kN

254 � 254 � 107 UC

6 mm fillet weld

M 20 class 4,6 bolts

Grade S355 steelFIGURE 7.28 ‘Pinned’column to foundation joint

Ch07-H5060.tex 12/7/2007 14: 23 page 246


Assuming that the bearing area is bounded by the dotted line shown previously inFig. 7.17 and the pressure beneath base plate (cl 6.2.5(7), EN 1993-1-8 (2005)),

NA

= fj = 23

fck

Rearranging and inserting known numerical values, bearing area required

A = NEd

fj= NEd(

23 fck

) = 1500E3(2×20

3

) = 112,5E3 mm2 (i)

To determine the minimum thickness of the steel base plate the bearing area enclosedby the dotted line (Fig. 7.17)

A = (bf + 2c)(h + 2c) − (h − 2tf − 2c)(bf − tw)

= (258,3 + 2c)(266,7 + 2c) − (266,7 − 2 × 20,5 − 2c)(258,3 − 13)

= 13524,4 + 1540,6c + 4c2 (ii)

Equating Eqs (i) and (ii) the projection of the dotted area

c = (192,572 + 24744)1/2 − 192,57 = 56,1 mm

Check if areas overlap along the bolt line between flanges

(h − 2tf )2

= (266,7 − 2 × 20,5)2

= 112,85 > (c = 56,1) mm satisfactory.

Thickness of base plate Grade S355 steel (cl 6.2.5(4), EN 1993-1-8(2005))

tp = c(

3fjγM0

fy

)1/2

= 56,1

(3 × 2

3 × 20 × 1355

)1/2

= 18,8 mm, use 20 mm thick base plate.

Minimum length of base plate

Dp = h + 2c = 266,7 + 2 × 56,1 = 322,8 mm

Minimum breadth of base plate

Bp = bf + 2c = 258,3 + 2 × 56,1 = 370,5 mm

Use 450 × 425 × 20 mm base plate Grade S355 steel.

If the end of the column is machined then the load is assumed to be transferred directlyto the base plate and a minimum size of fillet weld of 6 mm is used to connect the baseplate to the column.

Ch07-H5060.tex 12/7/2007 14: 23 page 247


Alternatively if the end of the column is not machined then the force per unit lengthof weld is approximately

Fw,Ed = NEd

(4bf+2h)= 1500

(4 × 258,3 + 2 × 266,7)= 0,957 kN/mm

Assuming a 6 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005))

Fw,Rd = fua(31/2βwγM2)

= 510 × 0,7 × 6/1E331/2 × 0,9 × 1,25)

= 1,1 > (Fw,Ed = 0,957) kN/mm satisfactor.

The base plate is subject to a compressive force which is not transferred to the holdingdown bolts. The bolts are therefore subject only to erection forces and if these are notknown then experience has shown that a bolt size approximately equal to the platethickness is suitable. Use 2M20 class 4.6 holding down bolts.

If there is a bending moment applied to the column and hence to the base plate thenthe bearing area is located beneath the column flange as shown in Fig. 6.4, EN 1993-1-8(2005).

EXAMPLE 7.7 ‘Rigid’ column bracket. Determine the size of fillet welds for thebracket shown in Fig. 7.29 at the ultimate limit-state.

End elevationSide elevation

Grade S 355 steel

h � 339,9

d � 246,7

Lg � 700

tw � 16,1

tf � 31,4

bf � 314,1 e � 600

V � 405 kN

N � 500 kN

bf � 293,8

Bg � 526

838 � 292�226 kg UB

305 � 305 � 198 kg UC

tf � 26,8

d t �

824

,1

Hg

� 7

97,3

t w �

19,

2r

� 1

5,2

r � 17,8

h �

850

,9

Ro

GG

oo

100

y

z

FIGURE 7.29 ‘Rigid’ column bracket

Ch07-H5060.tex 12/7/2007 14: 23 page 248


There are two possible solutions based on failure mechanisms (a) assuming rotationabout axis G–G which is the simple traditional conservative method and (b) assumingrotation about axis O–O which is more correct but the calculations are more extensive.

(a) Rotation about axis G–G

The fillet weld is continuous round the bracket section as shown in Fig. 7.29. If thereare no stiffeners in the web of the column then the strength of the weld around theflanges of the bracket is reduced because of the flexibility of the column flange.

Effective length of the column flange weld (Eq. (4.6a), EN 1993-1-8(2005))

beff = tw + 2r + 7ktf = 19,2 + 2 × 15,2 + 7 × 1 × 31,4 = 269,4 mm

which includes

k =(

tftp

)(fy,f

fy,p

)=(

31,426,8

)(355355

)= 1,17 > 1 use 1.

Check if stiffeners required for column web (Eq. (4.7), EN 1993-1-8(2005))

beff =(

fyp

fup

)bp = 355

510× 293,8

= 204,5 < (beff = 269,4) mm therefore no stiffeners required.

Rotation about axis G–G

The effective second moment of area of the weld group about axis G–G

IG = 2d3w

12+ 4beff (df/2)2

= 2(850,9 − 2 × 26,8)3

12+ 4 × 269,4 ×

[(850,9 − 26,8)

2

]2

= (84,5 + 183,0)E6 = 267,5E6 mm4

Maximum force per unit length on weld in the y direction (Eq. (7.28))

Fy =(Ve)

(df2

)IG

= 405 × 600 × (850,9 − 26,8)(2 × 267,5E6)

= 0,374 kN/mm

Maximum force per unit length of weld in the z direction

Fz = V/Lw = V[4beff + 2(h − 2tf )]

= 4054 × 269,4 + 2 × (850,9 − 2 × 26,8)

= 0,152 kN/mm

Maximum resultant design force per unit length of weld

Fr,Ed = (F2y + F2

z )1/2 = (0,3742 + 0,1522)1/2 = 0,404 kN/mm

Ch07-H5060.tex 12/7/2007 14: 23 page 249


For a 6 mm fillet weld for Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005))

Fw,Rd = fua

(31/2βwγM2)= 510 × 0,7 × 6/1E3

(31/2 × 0,9 × 1,25)

= 1,10 > (Fr,Ed = 0,404) kN/mm satisfactory.

Rotation about axis O–O

The second moment of area of the weld group about axis O–O

Io =(

23

)d3

w + 2beffd2f

=(

23

)(850,9 − 2 × 26,8)3 + 2 × 269,4 × (850,9 − 26,8)2

= (337,9 + 365,9)E6 = 703,8E6 mm4

Maximum force per unit length of weld in the y direction

Fy = (Ve)df/Io = (405 × 600) × (850,9 − 26,8)/703,8E6 = 0,285 kN/mm

Effective length of weld resisting shear

Leff = 4beff + 2dw = 4 × 269,4 + 2(850,9 − 2 × 26,8) = 2672 mm

Distance (dr) from the axis O–O to the resultant force in the weld is determined fromequating the moments of the forces in the weld group about the axis O–O

moment of the parts = moments of the whole

(2beff + 0,5 × 2df )Fxdr = FxIo/df

Rearranging and putting Io = (2/3)d3f + 2beffd2

f

dr

df=(

23 + 2beff

df

)(

1 + 2beffdf

)

=[

23 + 2×269,4

850,9−26,8

][1 + 2×269,4

850,9−26,8

] = 0,798

Check whether slip occurs by substituting in Eq. (7.31)

μsM(Vdr)

= μsedr

= 0,45 × 600[0,798(850,9 − 26,8)]

= 0,411 < 1, therefore slip occurs.

Maximum force per unit length of weld in the z direction

Fz = VLeff

− μsRLeff

= VLeff

− μs(Ve/dr)Leff

= 4052672

−0,45

[405×600

0,798(850,9−26,8)

]2672

= 0,089 kN/mm

Ch07-H5060.tex 12/7/2007 14: 23 page 250


Maximum resultant design force per unit length of weld

Fr,Ed = (F2y + F2

z )1/2 = (0,2852 + 0,0892)1/2 = 0,299 kN/mm

For a 6 mm fillet weld and Grade S355 steel (Eq. (4.4), EN 1993-1-8(2005))


= 510 × 0,7 × 6/1E3(31/2 × 0,9 × 1,25)

= 1,10 > (Fr,Ed = 0, 299) kN/mm satisfactory.

An alternative method related to the European Code:

Assume the applied vertical shear force is resisted by the two 6 mm web welds

VR,Ed = 2dfFw,Rd = 2 × (850,9 − 2 × 26,8) × 1,1

= 1450 > (VEd = 405) kN satisfactory,

and the applied bending moment is resisted by two 6 mm effective flange welds withrotation about axis O–O (cl 6.2.7.1(4), EN 1993-1-8(2005))

MR,Ed = 2beffdfFw,Rd = 2 × 269,4 × (850,9 − 26,8) × 1,1/1E3

= 488,4 > (MEd = 243) kNm satisfactory.

Gusset plate design

Check the thickness of the web of the 838 × 292 × 226 kg UB acting as a gusset plate.From Eq. (7.26)

Bg = Lg

[(Lg/Hg)2 + 1]1/2 = 700[(700/797,3)2 + 1]1/2 = 526,0 mm

Required thickness of the web of the UB acting as a gusset plate (Eq. (7.25))

tg = 2Pusg(fgyB2

g

γM1

) + Bg

80

= 2 × 405E3 × 600(355/1,0 × 526)2 + 526

80

= 11,52 < 16,1 mm (thickness of web of UB), satisfaactory.

Check the slenderness ratio of the web of UB acting as a gusset plate (Eq. (7.27))

lgig

= 2√

3Bg

tg= 2

√3 × 526

16,1

= 113,2 < 185 limit of application of the theory, acceptable.

Column web in transverse compression

Reaction Ro may buckle or crush the web of the 305 × 305 × 198 kg UC.

Ch07-H5060.tex 12/7/2007 14: 23 page 251


Reaction (Eq. (7.30))

Ro,Ed = Vedr

= 405 × 600[0,798(850,9 − 26,8)]

= 369,5 kN

The design resistance of the unstiffened column web (Eq. (6.9), EN 1993-1-8(2005))

Fc,wc,Rd = ωkwcbeff,c,wctwefywe

γM0

= 0,764 × 1 × 271,7 × 19,2 × 3551,0 × 1E3

= 1415 > (Ro,Ed = 369, 5) kN satisfactory.

or

Fc,wc,Rd = ωkwcρbeff,c,wctwefywe

γM1

= 0,764 × 1 × 1 × 271,7 × 19,2 × 3551,0 × 1E3

= 1415 > (Ro,Ed = 369,5) kN satisfactory.

From (Eq. (6.10), EN 1993-1-8(2005)) for a welded connection

beff,c,wc = tf b + 2 × 21/2ab + 5(tfc + s)

= 26,8 + 2 × 21/2 × 0,7 × 6 + 5 × (31,4 + 15,2) = 271,7 mm

The maximum longitudinal stress in the flange of the column from the axial andeccentric loads (Eq. (6.14), EN 1993-1-8(2005))

σcom,Ed = (N + V )A

+ V (e + h/2)Wel

= (500 + 405)E3252E2

+ 405E3(600 + 339, 9/2)2993E3

= 140,1 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

and (Table (6.3), EN 1993-1-8(2005))

β = 1(Table 5.4, EN 1993-1-8(2005))

ω = ω1 = 1[1 + 1,3

(beff,c,wctwc

Avc

)2]1/2

= 1[1 + 1,3 ×

(271,7 × 19,2

7032

)2]1/2 = 0,764

Ch07-H5060.tex 12/7/2007 14: 23 page 252


where the shear area of the column (cl 6.2.6(3), EN 1993-1-1(2005)) or from SectionTables

Avc = A − 2btf + (tw + 2r)tf

= 252E2 − 2 × 314,1 × 31,4 + (19,2 + 2 × 15,2) × 31,4 = 7032 mm2

ηhwtw = 1,0 × (339,9 − 2 × 31,4) × 19,2 = 5320 < (Avc = 7032) mm2

Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8(2005))

λp = 0,932[

beff,c,wcdwc fy,wc

(Et2wc)

]1/2

= 0,932 ×[

271,7 × 246,7 × 355(210E3 × 19,22)

]1/2

= 0,516 < 0,72 use ρ = 1,0

Shear strength of the column web (cl 6.2.6.1, EN 1993-1-8(2005))

Vwp,Rd = 0,9Avcfy,wc

(31/2γM0)= 0,9 × 7032 × 355

31/2 × 1 × 1E3

= 1297 > (Ro,Ed = 369,5) kN satisfactory.

EXAMPLE 7.8 ‘Rigid’ beam-to-column connection. Determine the size of thecomponents for the connection shown in Fig. 7.30 at the ultimate limit state.

tw � 7,7

d 1 �

339

,05

d 2 �

254

,05

d 3 �

54,

05

d f �

297

,2

bf � 166,8

t f �

13,

7

tw � 7,7tw � 13,0

tf � 20,5

r � 10,2

d � 160,8

h � 222,3

d �

265

,6

h �

310

,9

bf � 208,5

bf � 166,8

N � 100 kN

M �97,5 kNm

V � 60 kN

305�165�54 UB

203 � 203 � 86 UC

Grade S 355 steelM20 pre-loaded bolts

RoO O O

O O

oo

End plate 200 � 400 � 20 mm

End plate

90

2045

4585

200

70

35

359,

1

QbeFbt

(a) (b) (c)

FIGURE 7.30 ‘Rigid’ beam-to-column joint

Ch07-H5060.tex 12/7/2007 14: 23 page 253


Check for slip assuming rotation about axis O–O (Eq. (7.31))

μM[V (h − tf b)

= 0,45 × 97,5E6[60E3 × (310,9 − 13,7)]

= 2,46 > 1

therefore rotation about the compression flange of the beam at O without slip.

Assuming rotation about axis O–O the design tensile force acting on a single bolt

Ft,Ed = M[4(h − tf b)]

= 97,5E6[4 × (310,9 − 13,7) × 1E3]

= 82 kN

Design tensile resistance of an M20 pre-loaded class 8.8 bolt (Table 3.4, EN 1993-1-8(2005)) assuming not subject to a shear force

Ft,Rd = k2Asfub

γM2= 0,9 × 245 × 800

1,25 × 1E3

= 141 > (Ft,Ed = 82) kN satisfactory.

Shear resistance of an M20 class 8.8 pre-loaded bolt in single shear and subject to atensile force of 82 kN (Eq. (3.8a), EN 1993-1-8 (2005))

Fs,Rd = ksnμ

γM3(Fp,C − 0,8Ft,Ed)

= 1,0 × 1,0 × 0,51,25

× (0,7 × 800 × 245 − 0,8 × 82E3)1E3

= 28,6 >

(Fs,Ed = 60

6= 10

)kN satisfactory.

M20 class 8.8 bolt in bearing on end plate (t = 20 mm) (Table 3.4, EN 1993-1-8 (2005)).

Fb,Rd = k1αbfupdtγM2

= 2,5 × 0,682 × 510 × 20 × 201,25 × 1E3

= 278,2 > (Fs,Ed = 10) kN

for an end bolt αb = e1/(3do) = 45/(3 × 22) = 0, 682

for an edge bold k1 = 2,8 (e2/do) − 1,7 = 2,8 × (55/22) − 1,7 = 5,3 > 2,5

Thickness of end plate related to a single bolt (Table 6.2, EN 1993-1-8 (2005),Method 1)

tf =⎡⎢⎣4Ft,Edm( leff fy

γM0

)⎤⎥⎦

1/2

=[

4 × 82E3 × 35100 × 355/1,0

]1/2

= 18,0 mm; use 20 mm plate.

Ch07-H5060.tex 12/7/2007 14: 23 page 254


Design resistance of the unstiffened column web at O (Eq. (6.9), EN 1993-1-8 (2005))


γM0

= 0,685 × 1 × 224,1 × 13 × 3551,0 × 1E3


or


γM1

= 0,685 × 1 × 1 × 224,1 × 13 × 3551,0 × 1E3


where for a bolted end plate connection (Eq. (6.12), EN 1993-1-8 (2005))

beff,c,wc = tfb + 2 × 21/2ap + 5(tfc + s) + sp

= 13,7 + 2 × 21/2 × 0,7 × 6 + 5 × (20,5 + 15,2) + 20 = 224,1 mm

The maximum longitudinal stress in the flange of the column from the axial load andeccentric load (Eq. (6.14), EN 1993-1-8 (2005))

σcom,Ed = (N + V )A

+(Vh/2

)Wel

= (100 + 60)E3110E2

+60E3×223,2

2851E3

= 22,4 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

Table 5.4, EN 1993-1-8 (2005)

β = 1

Table 6.3, EN 1993-1-8 (2005)

ω = ω1 = 1[1 + 1,3

(beff,c,wctwc

Avc

)2]1/2

= 1[1 + 1,3 ×

(224,1 × 13

3124

)2]1/2 = 0,685

where (cl 6.2.6(3), EN 1993-1-1 (2005)) or from Section Tables


= 110E2 − 2 × 208,8 × 20,5 + (13 + 2 × 10,2) × 20,5 = 3124 mm2

ηhwtw = 1,0 × (222,3 − 2 × 20,5) × 13 = 2357 < (Avc = 3124) mm2

Ch07-H5060.tex 12/7/2007 14: 23 page 255


Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8 (2005))

λp = 0,932[

beff,c,wcdwefy,wc

(Et2wc)

]1/2

= 0,932[

224,1 × 161,8 × 355210E3 × 132

]1/2

= 0,561 < 0,72 use ρ = 1,0

Reaction Ro = 4Ft = 4 × 82 = 328 kN

Shear strength of the column web (cl 6.2.6.1, EN 1993-1-8 (2005))

Vwp,Rd = 0,9Avcfy,wc

(31/2γM0)= 0,9 × 3124 × 355

(31/2 × 1 × 1E3)

= 576 > (Ro = 328) kN satisfactory.

Alternative calculations for comparison

Previous calculations assume the top four bolts resist all of the applied moment witha single lever arm. Alternatively assume a linear variation of forces from axis O–O tothe bolts furthest from the axis.

Maximum tensile force acting on the bolt furthest from axis O–O

Ft = M[2(d1 + d2 + d3

)]= 97,5E3

[2(339,05 + 254,05 + 54,05)]= 75,3 kN

Prying force for each bolt assuming ∂b = 0 for a pre-loaded bolt and elastic behavior(Holmes and Martin, 1983).

Qbe =

[Fbe − 2EI∂b

apb2p

][

2ap

bp+(

13

) (ap

bp

)2]

= [90,62 − 0][2×45

35 +(

13

)×(

4535

)2] = 29,02 kN

and the maximum tensile force on a bolt

Fbt + Qbe = 75,3 + 29,0 = 104,3 < (Ft,Rd = 141) kN satisfactory for no slip.

For the welded connection between the end plate and the beam assume the appliedvertical shear force is resisted by two 6 mm web welds

VR,Ed = 2dfFw,Rd = 2 × (310,9 − 2 × 13,7) × 1,1

= 623,7 > (VEd = 60) kN satisfactory.

Ch07-H5060.tex 12/7/2007 14: 23 page 256


Assume that the applied bending moment is resisted by two 10 mm effective flangewelds with rotation about axis O–O (cl 6.2.7.1(4), EN 1993-1-8 (2005))

MR,Ed = 2beffdfFw,Rd = 2 × 91,2 × (310,9 − 13,7) × 1,831E3

= 99,2 > (MEd = 97,5) kNm satisfactory (cl 6.2.3(4), EN 1993-1-8 (2005)).

For strength of welds (Fw,Rd) see Annex 1 and for beff see Eq. (4.6a), EN 1993-1-8(2005).

EXAMPLE 7.9 ‘Rigid’ beam-to-beam joint. Determine the size of the componentsfor the rigid beam-to-beam joint shown in Fig. 7.31 at the ultimate limit state.

Assuming rotation about axis O–O and the applied moment of 97,5 kNm is resistedentirely by the cover plate then the design tensile force in the flange cover plate

Ff = Mh

= 97,5E6303,8 × 1E3

= 320,9 kN.

Thickness of flange connection plate, bp = 165 and 20 mm pre-loaded bolts

tp = Ff[(bp − 2dh)

(fy

γM1

)] = 320,9E3[(165 − 2 × 22) × 355

1,0

]= 7, 47 mm, use 8 mm thick Grade S355 steel plate.

Connecting plate


6 mmfillet weld

R0

305�165�40 kg UBbf � 165,1tw � 6,1

h �

465

,1

V � 45 kN

M � 97,5 kNm

h �

303

,8

tw � 10,7

tp � 10

t f �

10,

2

t p �

8

bf � 153,5

t f �

18,9O

10 mm thick end plate

457 � 152 � 82 kg UB

(a) Welded end plate connection

4060 6040

30

Alternative connections

(b) With bottomcleats

(c) With webcleats

75

FIGURE 7.31 ‘Rigid’ beam-to-beam joint

Ch07-H5060.tex 12/7/2007 14: 23 page 257


Single shear resistance of M20 class 8.8 pre-loaded bolt (Table 3.4, EN 1993-1-8 (2005))

Fs,Rd = ksnμ0,7Fp,C As

γM2

= 1,0 × 1, 0 × 0,5 × 0,7 × 800 × 2451E3

1,25= 54,9 kN

M20 class 8.8 bolt in bearing on plate (t = 8 mm) (Table 3.4, EN 1993-1-8 (2005)).

Fb,Rd = k1αb fupdtγM2

= 2,34 × 0,606 × 510 × 20 × 81,25 × 1E3

= 92,6 > (Fs,Rd = 54,9) kN

for an end bold αb = e1/(3do) = 40/(3 × 22) = 0,606

for an edge bolt k1 = 2,8(e2/do) − 1,7 = 2,8 × 31,75/22 − 1,7 = 2,34 < 2,5

Number of bolts required for the connecting plate

nb = Fs,Ed

Fs,Rd= 320,9

54,9= 5,85 use 6-M20 class 8.8 pre-loaded bolts.

Reaction at the hinge is equal to the force in the flange, Ro = Ff = 320,9 kN and thefrictional resistance at the hinge

μRo = 0,45 × 320,9

= 144,4 > (VEd = 45) kN therefore no slip occurs.

Use a 10 mm thick end plate welded to the end of the 305 × 165 × 40 kg UB and boltedto the 457 × 152 × 82 kg UB as shown in Fig. 7.31(a).

Shear resistance of 4-M20 pre-loaded bolts in double shear in the end plate

= 2 × 4 × 54,9 = 439,2 > (VEd = 90) kN satisfactory.

Shear resistance of two 6 mm fillet welds connecting the end plate to the305 × 165 × 40 kg UB using grade S355 steel

Fw,Rd = fua

(31/2βwγM2)

= 510 × 0,7 × 61E3

(31/2 × 0,9 × 1,25)= 1,10 kN/mm

2lwFw,Rd = 2 × (303,8 − 30 − 10,2) × 1,10

= 580 > (VEd = 45) kN satisfactory.

EXAMPLE 7.10 ‘Rigid’ beam splice. Determine the size of components for thebeam splice shown in Fig. 7.32 at the ultimate limit state.

Ch07-H5060.tex 12/7/2007 14: 23 page 258


V � 600 kN

M � 450 kNm40 40

280 80

Web plate10 mm thick

bf � 209,3

t f �

15,

6

h �

533

,1

140

88

3535

SectionGrade S355 steel

533 � 210 � 92 kg UBtw � 10,2

40 80 80

4040

80 V � 600 kN

M20 pre-loaded bolts

(b) Web connection resisting shear force and web bending moment

Mw � 87,4 kN

e � 121

8080

80

(a) Elevation z

y y

z

FIGURE 7.32 ‘Rigid’ beam splice

Check if the beam is in the elastic stage of behaviour

f = MWe

= 450E62076E3

= 216, 8 < ( fy = 355) MPa, therefore elastic behaviour.

Second moment of area of the web of the beam

Iweb = tw(h − 2tf )3

12= 10,2 × (533,1 − 2 × 15,6)3

12= 107, 5E6 mm4

From Section Tables the gross second moment of area of the beam section

Igross = 553,5E6 mm4

Assumed proportion of the applied bending moment taken by the web

Mweb =(

Iweb

Igross

)M =

(107,5E6553,5E6

)× 450 = 87,39 kNm

Check the strength of the arrangement of bolts in shear in the web plate (Fig. 7.32(b))

Ch07-H5060.tex 12/7/2007 14: 23 page 259


Second moment of area of bolts of unit area about the centroidal y–y axis

Iy = �(∂A)z2 = 6 × (802 + 1602) = 192E3 mm4

Second moment of area of bolts of unit area about the centroidal z–z axis

Iz = �(∂A)y2 = 10 × 802 = 64E3 mm4

Second moment of area of bolts of unit area about the centroidal polar x–x axis

Ix = Iy + Iz = (192 + 64)E3 = 256E3 mm4

Eccentricity of the applied shear force relative to the centroid of half the bolt groupis 121 mm (Fig. 7.32(b)). This eccentricity produces a moment which is increased bythe bending moment resisted by the web. The equivalent eccentricity

e′ = e + Mweb

V= 121 + 87,4E3

600= 266,6 mm

Maximum vector shear force in the y–y direction acting on a bolt furthest from thecentroid of the web bolt group

Fy = Ve′ zn

Ix= 600 × 266,6 × 160

256E3= 99,99 kN

Maximum vector shear force in the z–z direction acting on the same bolt

Fz = Vn

+ Ve′ yn

Ix= 600

15+ 600 × 266,6 × 80

256E3= 90,0 kN

Resultant maximum vector force acting on the same bolt

Fr = [F2y + F2

z ]1/2 = [99,992 + 90,02]1/2 = 134,5 kN

Double shear strength of an M20 pre-loaded bolt class 10,9 ( fpu = 1000 MPa, μ = 0,5)in the web (cl 3.9.1, EN 1993-1-8(2005))

Fv,Rd = ksnμFp,C

γM3= 1, 0 × 2, 0 × 0,5 × 245 × 0,7 × 1000

1E31,25

= 137,2 > (Fr = 134,5) kN, satisfactory.

M20 class 8.8 bolt in bearing on the web (t = 10,2 mm) (Table 3.4, EN 1993-1-8(2005)).

Fb,Rd = k1αbfupdtwγM2

= 2,5 × 0,606 × 510 × 20 × 10,21,25 × 1E3

= 126,1 < (Fv,Rd = 137,2) kN


for an edge bolt k1 = 2,8e2/do − 1,7 = 2,8 × 106,5/22 − 1,7 = 11,9 > 2,5

Ch07-H5060.tex 12/7/2007 14: 23 page 260


Required number of M20 pre-loaded bolts in bearing for the flange splice

nb = Ff

Fb,Rd=[

(M−Mweb)(h−tf )

]Fb,Rd

=[

(450−87,4)E3(533,1−15,6)

]126,1

= 5,56 use 6 bolts.

Reduction factor for length of lap (cl 3.8, EN 1993-1-8 (2005))

βLf = 1 − (Lj − 15d)200d

= 1 − (2 × 80 − 15 × 20)200 × 20

= 1,035 use 1,0

Thickness of the outer and inner flange cover plates

tp = Ff(bf−2dh+2wp−2dh)fy

γM1

= 700,7E3

(209,3 − 2 × 22 + 2 × 70 − 2 × 22) × 3551,0

= 7,55 mm

Use 8 mm plates Grade S355 steel as shown in Fig. 7.32.

EXAMPLE 7.11 ‘Rigid’ column splice. Determine the size of the components forthe rigid column splice shown in Fig. 7.33 at the ultimate limit state.

Where column sections are of the same serial size it is possible to connect them directlywith web and flange plates. The ends of the column are machined and will be in contact.Rotation will take place about an axis near the outer edge of the flange of the uppercolumn.

Thickness of the flange plate, from moments of forces about the axis of rotation

tp =(

M − Nhu2

)[(bp − 2dh)

( fyγM1

)hu

]

=[480E6 − 712,5E3×355,6

2

][(365 − 2 × 24)

(3551,0

)× 355,6

]= 8,83 mm, use 10 mm Grade S355 steel plate.

Single shear strength of an M22 pre-loaded bolt class 10,9 ( fpu = 1000 MPa, μ = 0,5)in the flange (cl 3.9.1, EN 1993-1-8 (2005))

Fv,Rd = ksnμFp,C

γM2

= 1,0 × 1,0 × 0,5 × 303 × 0,7 × 10001E3

1,25= 84,8 kN.

Ch07-H5060.tex 12/7/2007 14: 23 page 261


(c) Welded and plate column connection

(b) Column sections of different serial size

Angle

(a) Column sections of the same serial size

Grade S355 steelM22 pre-loaded bolts

356 � 368 � 202 kg UCbf � 374,4

hL � 374,7

tf � 27

tw � 16,8

tw � 10,7

tf � 17,5

hu � 355,6

N � 712,5 kN

M � 480 kNm

356 � 368 � 129 kg UCbf � 368,3

14080

140

40

4080

8080

8080

40

40150

Packing

960 � 365 � 10Flange plate

V � 150 kN

Web platethickness10 mm

Packing

Machinedplate

Flangeplate

Packing

Welded endplates

FIGURE 7.33 ‘Rigid’ splices in steel columns

M22 class 10,9 bolt in bearing on plate (t = 10 mm) (Table 3.4, EN 1993-1-8 (2005)).


= 2,5 × 0,555 × 510 × 22 × 101,25 × 1E3

= 124,5 > (Fv,Rd = 84,8)kN



Ch07-H5060.tex 12/7/2007 14: 23 page 262


Number of bolts required

nb =[M − Nhu

2

](Fv,Rdhu)

=[480E6 − 712,5E3×355,6

2

]84,8E3 × 355,6

= 11,7 use 12-M22 bolts.

Reduction factor for length of lap (cl 3.8, EN 1993-1-8 (2005))

βLf = 1 − (Lj − 15d)200d

= 1 − (5 × 80 − 15 × 22)200 × 22

= 0, 984. This factor does not affect the number of bolts required.

Where the ends of the column are machined and in contact the horizontal shear forceon the column is resisted by the friction force, in part or whole, at the point of contact,that is, at the axis of rotation.

Assuming machined surfaces μ = 0,15 the frictional resistance

= μ

(Mhu

+ N2

)

= 0,15 ×(

480E3355,6

+ 712,52

)= 255,9 > 150 kN (applied shear force).

Theoretically no shear connection required but in practice a web plate is generallyprovided to align the webs.

If the frictional resistance is ignored then the web splice is designed to resist the entireshear force as follows:

Second moments of area of two bolts of unit area on one side of the web connectionabout the centroidal axes are:

Iy = 0

Iz = 2 × 752 = 11,25E3 mm4

Ix = Iy + Iz = 11,25E3 mm4

Vector force on a bolt in the y–y direction

Fx = Vnb

= 1502

= 75 kN

Vector force on a bolt furthest from the centre of rotation in the z–z direction

Fz = (Ve)yn

lx= 150 × 40 × 75

11,25E3= 40 kN

Maximum vector shear force on the same bolt

Fr = (F2y + F2

z )1/2 = (752 + 402)1/2 = 85 < (Fb,Rd = 124,5) kN, acceptable.

Ch07-H5060.tex 12/7/2007 14: 23 page 263


(a)

(c)

(b)

70

eb � 70

C �

104

,411

3,1

Dp � 800

Bp

� 4

60

Lq � 278,9

2/3 � 20

Hg

� 3

00

xp � 83,6

Sg � 247,1

B g

Fg

h � 222,3

Dw �

300

z

z

y yG

o(d) Column-to-gusset plate welds

(e) Base plate-to-gusset plate welds

Dp � 800

G

Go

o

N � 198 kN

H � 49,5 kN

M � 264 kNm

203 � 203 � 86 UCbf � 208,8; h � 222,3Grade S355 steel

M30 class 4,6 bolts

Gusset platetg � 12,5

300

35

fc = 20

FIGURE 7.34 ‘Rigid’ column-to-foundation joint

EXAMPLE 7.12 ‘Rigid’ built-up column base connection. Determine the size of thecomponents for the connection shown in Fig. 7.34 at the ultimate limit state assumingfc = 20 MPa.

Tensile strength of an M30 class 4.6 holding down bolt (Table 3.4, EN 1993-1-8 (2005))

Ft,Rd = 0,9Asfub

γM2= 0,9 × 561 × 400

1,25 × 1E3= 161,6 kN

Distance required between holding down bolts (Eq. (7.39))

dp = M[(N2

)+ nFt,Rd

] = 264E3(1982 + 2 × 161,6

) = 625,3 mm

The size of the base plate is determined as follows:

Assume a bolt edge distance of approximately 2d = 2 × 30 = 60 mm use 70 mm.

Ch07-H5060.tex 12/7/2007 14: 23 page 264


Total length of base plate

Dp = dp + 2eb = 625,3 + 2 × 70 = 765,3 mm, use length of 800 mm

Minimum width of base plate

Bp = bf + 2tg + washer + 2 × welds + 2eb

= 208,8 + 2 × 12,5 + 66 + 2 × 10 + 2 × 70 = 459,8 mm. Use width of 460 mm.

The thickness of gusset plate(tg) and size of welds are assumed at this stage.

Assume the projection length for the base plate is half the width of the column(Fig. 7.34(b))

c = bf

2= 208,8

2= 104,4 mm

Thickness of base plate Grade S355 steel (Eq. (6.5), EN 1993-1-8 (2005))

tp = c(

3fjdγM0

fy

)1/2

= 104,4

(3 × 2

3 × 20 × 1,0

355

)1/2

= 35 mm

Length of concrete compression zone beneath the steel baseplate (Eq. (7.38))assuming lever arm la = 660 mm

xp =(

Mla

+ N2

)(bf + 2tg + 2c)fjd

=(

264E6660 + 198E3

2

)[(208,8 + 2 × 15 + 2 × 104,4) × 2

3 × 20]

= 83,6 mm

Lever arm for resistance of concrete in bending at ultimate limit state

la = Dp − eb − xp

2= 800 − 70 − 83,6

2= 688,2 mm

Tensile force in a holding down bolt (Eq. (7.39))

Fbt =M − N

(Dp

2 − Xp

2

)n la

b

=264E3 − 198 ×

(8002 − 83,6

2

)2 × 688,2

= 140,3 < (Ft,Rd = 161,6) kN

Use 800 × 460 × 35 mm base plate Grade S355 steel.

Force from bearing pressure applied to each gusset plate (Fig. 7.34(c))

Fg = fjd

(Bj

2

)xp = 2

3× 20 ×

(4602

)× 83,6

1E3= 256,4 kN

Ch07-H5060.tex 12/7/2007 14: 23 page 265


Length of gusset plate allowing for 10 mm for weld

Lg = (Dp − 2 × 10 − h)2

= (800 − 20 − 222,3)2

= 278,9 mm

Assume height of gusset plate Hg = 300 mm

Eccentricity of force Fg in relation to the inner corner of the gusset plate (Fig. 7.34(c))

sg = (Dp − h)2

− xp/2 = (800 − 222,3)2

− 83,6/2 = 247,1 mm

Width of gusset plate (Eq. (7.26))

Bg = Lg[1 +

( Lg

Hg

)2]1/2 = 278,9[

1 +(

278,9300

)2]1/2 = 204,2 mm

Thickness of gusset plate Grade S355 steel (Eq. (7.25))

tg = 2Fgsg( fyg

γM1 B2g

) + Bg

80

= 2 × 256,4E3 × 247,1(3551,0 × 204,22

) + 204,280

= 11,1 mm

Use 12,5 mm thick Grade S355 steel gusset plate.

Check slenderness ratio of gusset plate (Eq. (7.27))

lgig

= 2√

3Bg

tg= 2

√3 × 204,212,5

= 56,6 < 185, satisfactory.

Minimum length of foundation bolt (Holmes and Martin, 1983) (cl 6.2.6.12, EN1993-1-8 (2005))

Lb =√

Fbt

πftc=√√√√ 161,6E3

π × 0,3 × 202/3

1,5

= 186,8 mm

Use 4-M30 class 4.6 holding down bolts, 300 mm long anchored by washer plates.

The size of the fillet weld connecting the base plate to the gusset assuming no frictionis obtained as follows:

Length of weld (Fig. 7.34(e))

Lw = 2Dp = 2 × 800 = 1,6E3 mm

Second moment of area about centroid of the weld group for unit size weld

IwG = 2D3p

12= 2 × 8003

12= 85,33E6 mm4

Ch07-H5060.tex 12/7/2007 14: 23 page 266


Maximum vertical force compressive per unit length weld in the z direction

Fwz = NLw

+M(Dp

2

)Iwg

= 1981,6E3

+264E3 ×

(8002

)85,33E6

= 1,36 kN/mm

Horizontal force per unit length of weld in the x direction

Fwx = HLw

= 49,51,6E3

= 0,0309 KN/mm

Resultant vector force per unit length of weld

Fwr = (F2wz + F2

wx)1/2 = (1,362 + 0,03092)1/2 = 1,362 kN/mm


Fw,Rd = fua

(31/2βwγM2)= 510 × 0,7 × 8

1E3(31/2 × 0,9 × 1,25)

= 1,47 > (Fwr = 1,362) kN/mm

Alternatively if surface between the base plate and edge of the gusset plate aremachined and bearing is assumed at the axis of rotation O–O (Fig. 7.34(e)).

From Eq. (7.30)

Ro =⎛⎝ NDp

6 + M2Dp

3

⎞⎠ =

( 198×8006 + 264E3

2×8003

)

= 544,5 kN

Frictional resistance at Ro

= μRo = 0,15 × 544,5 = 81,65 > (H = 49,5) kN, satisfactory.

Second moment of area about axis O–O for unit size weld (Fig. 7.34(e))

IwO = 2D3p

3= 2 × 8003

3= 341,3E6 mm4

Maximum vertical force per unit length of weld in the z direction

Fwz = M − NDw2

IwO= 264E3 − 198×800

2341,3E3

= 0, 542 kN/mm



= 510 × 0,7 × 61E3

(31/2 × 0,9 × 1, 25)

= 1,10 > (Fwz = 0,542) kN/mm

Ch07-H5060.tex 12/7/2007 14: 23 page 267


If the end of the column is not machined then rotation is assumed to be about axis G–Gand the size of the weld connecting the column to the gusset plate is obtained as follows.

Length of weld (Fig. 7.34(d))

Lw = 4Dw = 4 × 300 = 1,2E3 mm

Second moments of area about the centroid of the weld group (axis G–G) fo unit sizewelds

IwGy = 4D3w

12= 4 × 3003

12= 9E6 mm4

IwGz = 4Dw

(h2

)2

= 4 × 300(

222,32

)2

= 14,83E6 mm4

Polar second moment of area

IwGx = IwGy + IwGz = (9,0 + 14,83)E6 = 23,83E6 mm4

Maximum vertical force per unit length of weld in the z direction on an element furthestfrom the axis of rotation

Fwz = NLw

+(

M − HDw

2

) (h2

)IwGx

= 1981,2E3

+(

264E3 − 49,5 × 3002

)×(

222,32

)23,83E6

= 1, 362 kN/mm

Horizontal force per unit length of weld in the y direction on the same element

Fwy = HLw

+(

M − HDw

2

) (Dw2

)IwGx

= 49,51,2E3

+(

264E3 − 49,5 × 3002

)×(

3002

)23,83E6

= 1,656 kN/mm


Fwr = (F2wy + F2

wz)1/2 = (1,6562 + 1,3622)1/2 = 2,144 kN/mm


Fw,Rd = fua

(301/2βwγM2)= 510 × 0, 7 × 12

1E3(31/2 × 0, 9 × 1, 25)

= 2, 20 > (Fwr = 2, 144) kN/mm

Alternatively if the contact between the base plate and the end of the column ismachined then the axis of rotation is O–O (Fig. 7.34(d)).

Ch07-H5060.tex 12/7/2007 14: 23 page 268


Second moments of area about centroid of the weld group for unit size weld aboutaxis O–O

IwOy = 4D3w

3= 4 × 3003

3= 36E6 mm4

Iwbz = 2Dwh2 = 2 × 300 × 222,32 = 29, 65E6 mm4

Polar second moment of area

IwOx = IwOy + IwOz = (36 + 29, 65)E6 = 65,65E6 mm4

Force per unit length of weld in the z direction on an element furthest from the axisof rotation

Fwz =(

M − Nh2

)h

IwOx

=(

264E3 − 198 × 222,32

)× 222,3

65,65E6= 0, 819 kN/mm

Force per unit length of weld in the y direction on the same element

Fwy =(

M − Nh2

)Dw

IwOx

=(

264E3 − 198 × 222,32

)× 300

65,65E6= 1,106 kN/mm


Fwr = (F2wy + F2

wz)1/2 = (1,1062 + 0,8192)1/2 = 1,376 kN/mm



= 510 × 0,7 × 81E3

(31/2 × 0,9 × 1,25)

= 1,47 > (Fwr = 1,376) kN/mm

Check shear resistance beneath the base plate (cl 6.2.2(8), Eq. (6.3), EN 1993-1-8(2005))

F1,v,Rd = 107,7 kN (see Annex A2)

αb = 0,44 − 0,0003fyb = 044 − 0,0003 × 240 = 0,368

F2,v,Rd = αbfubAs

γMb= 0,368 × 400 × 561

1,25= 66,1 < 107,7 kN

Fv,Rd = 0,2N + nbF1,v,Rd = 0,2 × 198 + 4 × 66,1

= 304 > (H = 49,5) kN satisfactory.

Ch07-H5060.tex 12/7/2007 14: 23 page 269


h1 � 250

N � 66,75 kN

M � 85 kNm

250 � 150 �10 RHS

300 � 200 �10 RHS

V � 40 kN 8 mm fillet weld

R0

o

Side wall

Side elevation

Chord face

250

150

o

o

R

dwr

Plan of weld

Assumedstress distribution

b1 � 150

t1 � 10

t0 � 10

b0 � 200

h 0 �

300

End elevation

FIGURE 7.35 ‘Rigid’ RHS joint

EXAMPLE 7.13 ‘Rigid’ RHS connection. Check the strength of the rigid rectan-gular hollow section connection at the ultimate limit state assuming Grade S355 steel(Fig. 7.35).

Check validity of the joint (Table 7.8, EN 1993-1-8 (2005))

b1

t1= 150

10= 15 < 35 satisfactory.

h1

t1= 250

10= 25 < 35 satisfactory.

Check for chord face failure of the horizontal member (Table 7.14, EN 1993-1-8(2005)).

Mip,1,Rd = knfyot2oh1

[1

(2η) + 2(1−β)1/2 + η

(1−β)

]γM5

= 1,0 × 355 × 102 × 250 ×

[1

(2×5/3)+ 2

(1−0,75)1/2 + 5/3(1−0,75)

](1,0 × 1E6)

= 97,3 > (MEd = 85) kNm satisfactory.

Ch07-H5060.tex 12/7/2007 14: 23 page 270


Included in the previous calculations

β = b1

b0= 150

200= 0,75 < 0,85

kn = 1,3 − 0,4n/β

= 1,3 − 0,4 × 0,2820,75

= 1,15 > 1 use 1

η = h1

b0= 250

150= 5

3

If the weld group is assumed to rotate about the axis O–O (Fig. 7.35) then the effectivewidth of the weld (Table 7.13, EN 1993-1-8 (2005)) furthest from the axis O–O

beff = 10fy0t0b1[(b0t0

)fy1t1

]

= 10 × 355 × 10 × 150[(20010

)× 355 × 10

] = 75 mm

beff

b1= 75

150= 0,5

Distance(dwr) from the axis O–O from the resultant force in the weld is obtained asfollows

moment of the parts = moment of the whole(beff + 1

2× 2dw

)Fzdwr = FzIoe

dwr

rearranging and substituting Ioe = 2d3w/3 + beffd2

w

dwr

dw=(

23 + beff

dw

)(

1+beffdw

)

=(

23 + 75

250

)(

1 + 75250

) = 0,744

Resultant reaction from Eq. (7.30)

Ro = M + N(dwr − dw/2)dwr

=[85E3 + 66,75 ×

(0,744 × 250 − 250

2

)]0,744 × 250

= 478,9 kN

Frictional force at the stiff bearing if the end of the vertical RHS is machined

μRo = 0, 15 × 478, 9 = 71, 8 > 40 kN (applied shear force).

The weld group is subject to the actions from N and M acting about axis O–O.

Ch07-H5060.tex 12/7/2007 14: 23 page 271


Second moment of area about axis O–O of the weld group (Fig. 7.35) for unit sizewelds

IO = 2d3w

3+ beffd2

w = 2 × 2503

3+ 75 × 2502 = 15,1E6 mm4

Moment applied about axis O–O

M ′ = M − Ndw

2= 85 − 66,75 × 250

2 × 1E3= 76,65 kNm

Maximum tensile force per unit length of weld furthest from the axis of rotation

Fw = M′d

IO= 76,65E3 × 250

15,1E6= 1,27 kN/mm


Fw,Rd = fua

(31/2βwγM2)= 510 × 0,7 × 8

1E3(31/2 × 0,9 × 1,25)

= 1,47 > (Fwy = 1,27) kN/mm

An alternative more conservative calculation to determine the size of weld is to assumerotation about the centroid of the weld group. This results in a larger weld size.

For interest check for side wall crushing in horizontal member at O–O (Table 7.14,EN 1993-1-8 (2005))

Mip,1,Rd = 0,5fykt0(h1 + 5t0)2

γM5

= 0,5 × 355 × 10 × (300 + 5 × 10)2

(1,0 × 1E6)= 217,4 > Mip,1,Rd = 155 kNm.

This calculation is not necessary because the ratio β = b1/b0 = 150/200 = 0,75 < 0,85indicates that it is not critical.

EXAMPLE 7.14 ‘Rigid’ knee connection for a portal frame. Check the strength ofthe knee joint components for the frame (Fig. 7.36) at the ultimate limit state.

Moment acting about axis O–O

MO = 250,2 − 116,8 × 0,5379 = 187,4 kNm

Thickness of strap assuming 190 mm wide and 22 mm diameter hole

tst = MO( labstfy,st

γM1

)= 187,4E6[

717,2 × cos 10◦ × (190 − 2 × 22) × 3551,0

]= 5,12 mm. Use a 8 mm thick strap.

Ch07-H5060.tex 12/7/2007 14: 23 page 272


tw � 8,5r � 10,2

457�191�67 kg UBbf � 189,9, h � 453,6

d � 407,9


6 mm fillet welds

Strap welded tocolumn t � 8

Web stiffenerts � 8

V � 174,4 kN

100 5013

x

10°

H �116,8 kN

tw � 8

537,

9

717,

2

M � 250,2 kNm

Separate bolted strap

Cap platet � 12,5

O

tf � 12,7

R0end plate t � 12,5

(a) (b)

FIGURE 7.36 ‘Rigid’ knee joint for a portal frame

Strap welded to the top of the column and bolted to the rafter (Fig. 7.36(a)).

For a single M20 pre-loaded bolt slip resistance (Eq. (3.6), EN 1993-1-8 (2005))

Fs,Rd = ksn μfubAs

γM3

= 1,0 × 1,0 × 0,4 × 0,7 × 800 × 2451,25 × 1E3

= 43,9 kN

M20 class 8.8 bolt in bearing on the plate (t = 8 mm) (Table 3.4, EN 1993-1-8 (2005)).


= 2,5 × 0,758 × 510 × 20 × 81,25 × 1E3

= 123,7 > (Fs,Rd = 43,9) kN



Number of M20 pre-loaded bolts in single shear for the strap

nb = MO

(laFs,Rd)= 187,4E3

717,2 × cos 10◦ × 43,9= 6,04 use 6 bolts.

Force per unit length acting on the two welds connecting strap to head of column

Fw,Ed = MO

lanwlw

= 187,4E3

717,2 × cos 10◦ × 2 × 407,9cos 10◦

= 0,320 kN/mm

Ch07-H5060.tex 12/7/2007 14: 23 page 273


Shear resistance of a 6 mm fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005))

Fw,Rd = fua

(31/2βwγM2)

= 510 × 0,7 × 6(31/2 × 0,9 × 1,25 × 1E3)

= 1,1 > (Fw,Ed = 0,320) kN/mm

These calculations assume, conservatively, that the force is resisted only by the twowelds along the web of the column, but the strap is also welded to the flanges of thecolumn which increases the strength.

Moments of forces about X to determine the reaction Ro at O

−717, 2Ro + 250, 2E3 + 116, 8 × (717, 2 − 537, 9) = 0; hence Ro = 378, 1 kN

Vertical frictional force at O

= μRo = 0, 4 × 378, 1 = 151, 2 < (VEd = 174, 4) kN, therefore slip occurs.

Bolt end plate (tep = 12,5 mm) to the flange of the column using M20 pre-loaded bolts(fpu = 800 MPa, μ = 0,4). For a single bolt slip resistance (Eq. (3.6), EN 1993-1-8(2005)). Fs,Rd = 43,9 kN as calculated previously.

Number of bolts required

nb = VEd

Fv,Rd= 174,4

43,9= 3,97 use 4 bolts.

Shear strength of the web of the column (cl 6.2.6.1 (2), EN 1993-1-8 (2005))

Vpl,Rd = 0,9Avc

( fy,wc

γM0

)31/2 = 0,9 × 4094 ×

(3551,0

)(31/2 × 1E3)

= 755,2 > (Ro = 378,1) kN, satisfactory.

Design resistance of the unstiffened column web at O (Eq. (6.9), EN 1993-1-8(2005))


γM0

= 0,932 × 1 × 0,731 × 164,4 × 8,5 × 3551,0 × 1E3

= 338,0 < (Ro = 378,1) kN not satisfactory.

or


γM1

= 0,932 × 1 × 0,731 × 164,4 × 8,5 × 3551,00 × 1E3

= 338,0 < (Ro = 378,1) kN not satisfactory stiffeners required.

Ch07-H5060.tex 12/7/2007 14: 23 page 274


Previous calculations include the effective width (Eq. (6.11), EN 1993-1-8 (2005))

beff,c,wc = tfb + 2 × 21/2ap + 5(tfc + s) + sp

= 13 + 2 × 21/2 × 0,7 × 6 + 5 × (12,7 + 10,2) + 2 × 12,5 = 164,4 mm

Maximum longitudinal stress in the flange of the column from the axial load andeccentric load (cl 6.2.6.2(2), EN 1993-1-8 (2005))

σcom,Ed = VA

+(

M + Vh2

)Wel

= 174,4E385,5E2

+(

250,2E6 + 174,4E3 × 453,62

)1297E3

= 244 < (0,7fy,wc = 0,7 × 355 = 248,5) MPa hence kwc = 1

β = 1 (Table 5.4, EN 1993-1-8 (2005)) and from Table 6.3, EN 1993-1-8 (2005)

ω = ω1 = 1[1 + 1,3

(beff,c,wctwc

Avc

)2]1/2

= 1[1 + 1,3

(164,4×8,5

4094

)2]1/2 = 0,932

which includes (cl 6.2.6(3), EN 1993-1-1 (2005)) or from Section Tables


= 85, 5E2 − 2 × 189, 9 × 12, 7 + (8, 5 + 2 × 10, 2) × 12, 7 = 4094 mm2

ηhwctwc = 1,0 × (453,6 − 2 × 12,7) × 8,5

= 3640 < (Avc = 4094) mm2 satisfactory.

Plate slenderness (cl 6.2.6.2(1), EN 1993-1-8 (2005))

λp = 0,932[

beff,c,wcdwefy,wc

(E t2wc)

]1/2

= 0,932[

164,4 × 407,9 × 355(210E3 × 8,52)

]1/2

= 1,167 > 0,72 therefore

ρ = λp − 0,2

λ2p

= 1,196 − 0,21,1962 = 0,731

Ch07-H5060.tex 12/7/2007 14: 23 page 275


Thickness of the load bearing web stiffeners in compression in the column (cl 6.2.6.2(4),EN 1993-1-8 (2005))

ts = Ro[(bcf − tcw − 2rc)

fyw

γM1

]= 378, 1E3[

(189,9 − 8,5 − 2 × 10,2) 3551,0

]= 6,62 mm. Use 8 mm thick stiffeners either side of the web of the column.

Force per unit length of weld connecting the web stiffener to the column web

Fw = Ro(nwlw

) = 378,12 × 407,9

= 0,463 kN/mm

As previously the shear resistance of a 6 mm fillet weld (cl 4.5.3.3, EN 1993-1-8 (2005))

Fw,Rd = 1, 1 > (Fw = 0,463) kN/mm

It is assumed conservatively that the force is resisted by the welds along the web.

To avoid damage to the strap in transit an alternative arrangement is shown inFig. 7.36(b).

7.11 JOINT ROTATIONAL STIFFNESS (CL 6.3, EN 1993-1-8 (2005))Previous sections in this chapter show calculations for joint strength. Another aspectof joint behaviour that needs consideration is the relationship between joint rota-tional stiffness and member stiffness. This relationship is important in the analysis ofstructures (cl 5, EN 1993-1-8 (2005)).

A simple example of joint rotational stiffness (Fig. 7.37(a)) is as follows. The theoreticalrotational stiffness related to the elastic and plastic behaviour of the top cleat can bedeveloped for the joint. Initially assuming elastic behaviour of the steel top cleat withdeformations as shown

Sj = Mφ

= Ftz(�

z

) =2(

6 E la�m3

)z2

�= 12E

(bat3

a

12m3

)z2

= E z2[1

(bat3a/m3)

] (7.40)

This equation considers only deformation of the top cleat and can be compared withthe theoretical expression (Eq. (6.27), EN 1993-1-8 (2005)) which replaces ba with leff

and includes factors for other deformations (Table 6.11, EN 1993-1-8 (2005)).

Sj = E z2[μ�(

1k1

)] (7.41)

Ch07-H5060.tex 12/7/2007 14: 23 page 276


M

(a)

(b) (c)

�

6E1� /m2

m

Z

A Mj

f

u � Mj/Sj

Q

L

0,4

0,3

0,2

0,1

0 0,2 0,4 0,6 0,8 1,0

Mj

Sj /kb

Q L/8

Hu

HI

2,0

0,5kc/kb

FIGURE 7.37 Stiffness of joints

Other factors that affect the rotational stiffness of a joint are: extension and sheardeformation of bolts, and deformation of column flanges, end plates and columnwebs. Each factor (Tables 6.10 and 6.11, EN 1993-1-8 (2005)) produces values of 1/k1

which are added and inserted in Eq. (7.41) as shown in Example 7.15.

Experimental results (Maxwell et al., 1981) which show that the rotational stiffness of ajoint is non-linear as ultimate resistance approaches. The stiffness of the joint reducesand to allow for this the value of μ increases (cl 6.3.1(6), EN 1993-1-8 (2005)).

It is also necessary to check for other forms of failure. For this joint (Fig. 7.37(a)) theultimate moment of resistance based on yielding of the top angle

Mjya,Rd = 2

(b t2

a4

)(fy

mz

)(7.42a)

Ch07-H5060.tex 12/7/2007 14: 23 page 277


Alternatively for this joint if the bolts reach ultimate load

Mjub,Rd = nbAbfuz (7.42b)

7.12 FRAME-TO-JOINT STIFFNESS

Ideally the analysis of a structure should incorporate the stiffness of the joints and themembers. An example of the theoretical relationship between the stiffness of jointand members for a simple frame follows.

Consider the elastic behaviour of the simple frame in Fig. 7.37(b). From the applicationof the area moment method (Croxton and Martin, 1987 and 1989), the moment ofresistance of the joint (Mj) for a point load at the mid-span of the beam

For the beam at A

EIb

(θ + Mj

Sj

)= 1

2× Lb

2× QLb

4− MjLb

2(i)

For the upper column at A

EIcuθ = 14

× MjuHu (ii)

For the lower column at A

EIclθ = 14

× MjlHl (iii)

and

Mj = Mju + Mjl (iv)

Combining (i) to (iv) to eliminate θ and expressing stiffness as k = El/L

Mj =(

Q L8

)1 + 1

2(kcU/kb+kcL/kb)+ 2kb

Sj

(7.43)

The stiffness of the joint (Sj), beam (kb) and columns (kc) can be seen to affect themoment of resistance at the joint (Mj).

EXAMPLE 7.15 Stiffness of a ‘pin’ joint. Determine the stiffners of the ‘Pin’ shownin Fig. 7.26.

Joint stiffness (Eq. (6.27), EN 1993-1-8(2005)) assuming μ = 1

Sj = E z2[μ �

(1k

)]

= 210E3 × 517,42

[(1 × 2,48 × 1E6)]= 22668 kNm/radian

Ch07-H5060.tex 12/7/2007 14: 23 page 278


where (Fig. 6.15(b), EN 1993-1-8 (2005))

z = h + ea + ta2

= 467,4 + 45 + 102

= 517,4 mm

and for a column web panel in shear (Table 6.11, EN 1993-1-8 (2005))

k1 = 0,38Avc

β z= 0,38 × 3130

1 × 517,4= 2,3

and for a column web panel in compression (Table 6.11, EN 1993-1-8 (2005))

k2 = 0,7beff,c,wctwc

dc= 0,7 × 179,5 × 13

160,8= 10,2

where (Table 6.12, EN 1993-1-8 (2005))

beff,c,we = 2ta + 0,6ra + 5(tfc + s)

= 2 × 10 + 0,6 × 10 + 5 × (20,5 + 10,2) = 179,5 mm

and for a column web in tension (Table 6.11, EN 1993-1-8 (2005))

k3 = 0,7beff,c,wctfcdc

= 0,7 × 210 × 13160,8

= 11,9

where (Table 6.4, EN 1993-1-8 (2005)) beff,c,wc is the lesser value of

2πm = 2π × 45 = 283 mm and

πm + 2e1 = π × 45 + 2 × (208,8 − 140)2

= 210 mm

and for a column flange in bending (Table 6.11, EN 1993-1-8 (2005))

k4 = 0,9leff t3fe

m3 = 0,9 × 146,5 × 20,53

453 = 12,5

where (Table 6.4, EN 1993-1-8 (2005)) leff is the lesser value of

leff = 4m + 1,25e = 4 × 45 + 1,25 × (208,8 − 140)2

= 223 mm or

leff = 2m + 0,625e + e1

= 2 × 45 + 0,625 + (208,8 − 140)2

+ 35 = 146,5 mm

and for a cleat in bending (Table 6.11, EN 1993-1-8 (2005))

k6 = 0,9leff t3

m3 = 0,9(208,8)2

× 103(45 − 10/2

)3 = 1,03

and for bolts in tension (Table 6.11, EN 1993-1-8 (2005))

k10 = 1,6As

Lb= 1,6 × 245(

20,5 + 10 + 3,7 + 132 + 16

) = 6,91

Ch07-H5060.tex 12/7/2007 14: 23 page 279


and for bolts in shear (Table 6.11, EN 1993-1-8 (2005))

k11 = 16 nbd2fub

E dM16= 16 × 2 × 202 × 800

210E3 × 14,1= 3,45

and for bolts in bearing (Table 6.11, EN 1993-1-8 (2005))

k12 = 24 nbkbktd fuE

= 24 × 2 × 1,06 × 1,06 × 20 × 510210E3

= 2,62

where

kb = 0,25eb

d+ 0,5 = 0,25 × 45

20+ 0,5 = 1,06

kt = 1,5tjdM16

= 1,5 × 1014,1

= 1,06

The total value of

μ�

(1k

)= 1 ×

(1

2,3+ 1

10,2+ 1

11,9+ 1

12,5+ 1

1,03+ 1

6,91+ 1

3,45+ 1

2,62

)= 2,48.

EXAMPLE 7.16 Effect of ‘pin’ joint stiffness on a simple frame shown inFig. 7.37(b). Assume Q = 52 kN, Lb = 10 m, kb = 9612 kNm, kb/kc = 1 and Sj =22668 kNm obtained in Example 7.15.

The joint design moment of resistance at A (Eq. (7.43))

Mj,Rd =(

QL8

)⎛⎝1 + 1

2(

kcukb

+ kclkb

) + 2kbSj

⎞⎠

=52E3×10E3

(8×1E6)(1 + 1

2(1+1) + 2×961222668

)= 31 kNm

The effect of introducing the joint stiffness to the above equation is to reduce the endmoment on the beam (Mj,Rd) from 51,8 to 31 kNm.

If Mj,Ed = 20 kNm then

Mj,Ed/Mj,Rd = 20/31 = 0,65 < 2/3 and the adoption of μ = 1 is satisfactory (cl 6.3.1(6),EN 1993-1-8 (2005)).

Check the ultimate moment of resistance of the joint based on yielding of the top angle(Eq. (7.42a))

Mjya,Rd = 2

(b t2

a4

)(fy

m z

)

= 2 ×(

208,8 × 102

4

)×(

35540 × 517,4

)1E6

= 47,94 > (Mj,Rd = 31) kNm

Ch07-H5060.tex 12/7/2007 14: 23 page 280


which is the maximum moment of resistance for the joint.

Check the ultimate moment of resistance of the two top bolts (2M-20 class 8.8) failingin tension (Eq. (7.42b))

Mjub,Rd = 2Abfuz = 2 × 245 × 800 × 517,41E6

= 202,8 > (Mjya,Rd = 47,94) kNm

The effect of varying joint to beam stiffness can be shown for the simple structure(Fig. 7.37(b)) using Eq. (7.43). The ratio (Mj/(QL/8)), which includes the momentof resistance at the end of the beam, varies with the ratios of joint-to-beam stiffness(Sj/kb) and column-to-beam stiffness(kc/kb) as shown in Fig. 7.37(c). This relationshipis related to the structure in Fig. 7.37(b) for a beam stiffness of 4E3 kNm. Elasticbehaviour is assumed but plastic failure may limit the value of the joint resistance.

For example the joint moment (Mj) reduces the moment at mid-span for the beam,which is beneficial, but introduces a moment to the columns which if slender can reducetheir capacity. However for more complicated structures an increase in connectionstiffness reduces slenderness ratios and column deflections, which increase the loadcapacity of a column. Numerical investigations (Jones et al., 1981) show that includingjoint stiffnesses in the analysis of frames can reduce the weight of steel.

REFERENCES

Bahia, C.S. and Martin, L.H. (1980). Bolt groups subject to torsion and shear, Proceedings of theI.C.E. Pt2, V69.

Bahia, C.S. and Martin, L.H. (1981). Experiments on stressed and unstressed bolt groups subject totorsion and shear, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic.

Bahia, C.S., Graham, J. and Martin, L.H. (1981). Experiments on rigid beam-to-column connectionssubject to shear and bending forces, Conference Proceedings, Joints in Structural Steelwork. TeesidePolytechnic.

BS 7668 (1994), BSEN 10029 (1991), BSEN 10113-1 to 3 (1993) and BSEN 10210-1 (1994).Specification for weldable structural steels. BSI.

BS 5400 (2000). Steel concrete and composite bridges, Pt3 Code of practice for the design of steelbridges. BSI.

BS 3692 (2001). ISO Metric precision hexagon bolts, screws and nuts. BSI.BS 4190 (2001). ISO Metric black hexagon bolts, screws and nuts. BSI.BSEN 499 (1995). Covered electrodes for the manual metal arc welding of carbon and carbon manganese

steels. BSI.BSEN ISO 4320 (1998). Metal washers for general engineering purposes. BSI.BSEN 1011-1 (1998) and 2 (2001). Specification for the process of arc welding of carbon and carbon

manganese steels. BSI.BSEN 14399-1 to 5 (2005). High strength friction grip bolts and associated nuts and washers for structural

engineering: Pt1 General grade, Pt2 Higher grade bolts and nuts and general grade washers. BSI.Biggs, M.S.A.B., Crofts, M.R., Higgs, J.D., Martin, L.H. and Tzogius, A. (1981). Failure of fillet

welded connections subject to static loading, Conference Proceedings, Joints in Steelwork, TeesidePolytechnic.

Chesson, Jr., E. Faustino, N.L. and Munse, W.H. (1965). High strength bolts subject to torsion andshear, A.S.C.E. (Structural Division), V91, ST5.

Ch07-H5060.tex 12/7/2007 14: 23 page 281


Clarke, A. (1970). The strength of fillet welded connections. MSc Thesis, Imperial College, Universityof London.

Clarke, P.J. (1971). Basis for the design of fillet welded joints under static loading, ConferenceProceedings. Welding Institution, Improving Welding Design Paper 10, V1.

Croxton, P.C.L. and Martin, L.H. (1987 and 1989). Solving Problems in Structures Vols. 1 and 2.Longman Scientific and Technical.

Davies, G. (1981). Estimating the strength of some welded lap joints formed from rectangular hollowsection members, Conference Proceedings, Joints in Structural Steelwork, Teeside Polytechnic.

Elzen, L.W.A. (1966). Welding beams in beam-to-column connections without the use of stiffening plates.Report 6-66-2. I.I.W. document XV-213-66.

EN 1993-1-1 (2005). General rules and rules for buildings. BSI.EN 1993-1-8 (2005). Design of joints. BSI.European Convention for Structural Steelwork (1981). European recommendations for steel construc-

tion. Construction Press.Farrar, J.C.M. and Dolby, R.E. (1972). Lamellar tearing in welded steel fabrication. Welding Institute,

Cambridge, England.Fisher, J.W. and Struik, J.H.A. (1974). Guide to Design Criteria for Bolted and Riveted Joints. John

Wiley and Sons.Gourd, L.M. (1980) Principles of Welding Technology. Edward Arnold.Holmes, M. and Martin, L.H. (1983). Analysis and Design of Structural Connections. Ellis Hor-

wood Ltd.Jones, S.W., Kirby, P.A. and Nethercot, D.A. (1981). Modelling of semi-rigid connection behaviour

and its influence on steel column behaviour, Conference Proceedings, Joints in Steelwork, TeesidePolytechnic.

Kato, B. and Morita, K. (1974). Strength of transverse fillet welded joints. Welding Research.Ligtenberg, F.K. (1968). International test series, final report, Stevin Laboratory, Technological

University of Delft, Doc XV-242-68.Mann, A.P. and Morris, L.J. (1979). Limit state design of extended plate connections, ASCE Journal

of Structural Engineering, 105(ST3): 511–526.Martin, L.H. (1979). Methods for limit state design of triangular steel gusset plates, Building and

Environment.Martin, L.H. and Robinson, S. (1981). Experiments to investigate parameters associated with the

failure of gusset plates, Conference Proceedings, Joints in Steelwork, Teeside Polytechnic.Maxwell, S.M., Jenkins, W.M. and Howlett, J.H. (1981). A theoretical approach to the analysis of

connection behaviour. Conference Proceedings, Joints in Steelwork. Teeside Polytechnic.Morris, L.J. and Newsome, C.P. (1981). Bolted corner connection subjected to an out of balance

moment – the behaviour of the web panel, Conference Proceedings, Joints in Structural Steelwork,Teeside Polytechnic.

Owens, G.W., Driver, P.J. and Kriege, G.J. (1981). Punched holes in structural steelwork,Constructional Steel Research, 1(3).

Pillinger, A.H. (1988). Structural steel work: a flexible approach to the design of simple construction,Structural Eng 66(19/4), October.

Purkiss, J.A. and Croxton, P.C.L. (1981). Design of eccentric welded connections in rolled hollowsections, Conference Proceedings, Joints in Structural Steelwork. Teeside Polytechnic.

Rolloos, A. (1969). The effective weld length of beam-to-column connections with stiffening plates, Finalreport, I.I.W. Document XV-276-69.

Stamenkovic, A. and Sparrow, K.D. (1981). A review of existing methods for the determination of thestatic axial strength of welded T, Y, N, K, and X joints in circular hollow steel sections, ConferenceProceedings, Joints in Structural Steelwork. Teeside Polytechnic.

Stark, J.W.B. and Bijlaard, F.S.K. (1988). Design rules for beam-to-column connections in Europe, SteelBeam-to Column Building Connections. pub. Elsevier Applied Science.

Ch08-H5060.tex 11/7/2007 14: 11 page 282

C h a p t e r 8 / Frames and Framing

The previous chapters have dealt with design of beams, columns and connections. Thischapter, and the next, deal with the way the individual components are assembledtogether and also with design problems associated with the whole structure.

In the first place it is necessary to give some consideration to how the choice is madeof the structural form employed to carry the primary loading. For convenience thissurvey is divided into single and multi-storey structures.

8.1 SINGLE STOREY STRUCTURES

Typical examples of such structures include sports complexes, exhibition halls, factoryunits or assembly buildings. Unless architectural considerations prevail, the most eco-nomic solution will be obtained using one-way spanning structural systems rather thanspace frame structures. Systems which at first sight appear two-way spanning oftencomprise a number of overlaid one-way systems. Roof systems can be convenientlydivided into flat and pitched roof systems.

8.1.1 Flat Roof Systems

For spans up to around 15 m rolled sections form the most economic solution. How-ever, it should be noted that the potential extra cost of beams over 12 m long needsto be taken into consideration, and the use of 12–15 m long beams should only becontemplated if a large number are required. At around 14 m and up to 20 m the useof castellated or beams with circular openings in the web become economic. Althoughthis type of beam incurs high fabrication costs and requires a higher construction depththan ordinary rolled sections, the holes in the web allow services to be contained withinthe beam depth. Above 20 m it is usual to use parallel chord lattice trusses fabricatedeither from rolled hollow sections or from lightweight cold formed sections (top andbottom chords) and bar members for the web. The roof decking is then generallylightweight steel sheeting with suitable finishes and insulation, although timber deck-ing with asphaltic waterproofing, woodwool slabs or pre-cast concrete decking unitsmay be used. One problem with flat roofs is drainage and thus sufficient cross-fallmust be provided to give adequate run-off and avoid local ponding. The cross-fall is

Ch08-H5060.tex 11/7/2007 14: 11 page 283


generally provided by variable depth purlins and joists or by adjusting screed depths.To some extent these problems are alleviated by employing a sloping top chord to theroof system.

8.1.2 Pitched Roof Systems

These fall into two main divisions: trusses with a sloping top chord and pitched roofportal frames.

Modern trusses have a relatively low slope to the top chord. A pitch of between4◦ and 10◦ is adequate to allow run off and to allow the joints in the roof sheeting toremain watertight. This type of truss is generally fabricated from square or rectangularrolled hollow sections with fully welded nodal connections, with circular hollow sec-tions sometimes being used for the web members in Grade S355 steel, owing to avail-ability, Pitches greater than 10◦ are only seen where an existing building with traditionallarge pitch trusses designed for tiles or slates as roofing materials is being extended, orwhere a large pitch is required for architectural reasons in, for example, shopping malls.

The most common method of single storey construction is the pitched roof portal framewhether for factory units, small sports complexes or warehouses. This is basically dueto the high speed and simplicity of construction. The internal bays are designed asrigid jointed frames. The end frames, unless there is likely to be an extension to thestructure, are much lighter and have the rafters designed as spanning across the gableend posts which are also used to support the sheeting rails for the cladding. The mosteconomic frame spacing is generally 7.5 m or 9.0 m for much higher frame spans. Forspans below 20 m a frame spacing of 6.0 m may be adopted (Horridge, 1985; Horridgeand Morris, 1986).

Both truss systems and portal frames can be used for spans up to 60 m, althoughfor spans over 30 m multi-bay structures become an option. Multi-bay constructionrequires internal columns which may reduce the flexibility of usage. However, thissituation may be mitigated by the use of internal lattice girder support systems. Notethat where the roof system comprises trusses the internal support system can be withinthe depth of the truss. This is not possible with portal frames, so unless reducedheadroom is acceptable internal columns must be used. It should also be rememberedthat where multi-bay construction is used, there will be the need to supply valleyguttering and associated drainage. Such valleys cause build-up in snow loading andcan be potential areas of leakage and cause problems with access for maintenance.

8.2 MULTI-STOREY CONSTRUCTION

8.2.1 Multi-storey Steel Skeleton

In UK practice the steel skeleton is generally designed to carry vertical loading due tothe permanent and variable actions only, with the horizontal loading from wind and

Ch08-H5060.tex 11/7/2007 14: 11 page 284

284 • Chapter 8 / Frames and Framing

the notional horizontal loading taken by a bracing system, or more commonly the liftshaft(s) and stair well(s). The economics of various types of multi-storey construction isdiscussed in Gibbons (1995). When using lift shafts or stairwells as bracing care mustbe taken in their layout as torsional effects from lateral loading on an asymmetriclayout must be avoided.

8.2.2 Flooring Systems

The flooring system is generally required to act as a horizontal diaphragm to carryhorizontal forces from their point of application to the bracing or lift core. Thus it isessential that adequate lateral stiffness exists in the plane of the floor and the flooringsystem is adequately tied to the frame.

Historically the floors of multi-storey steel frames have comprised cast in-situ normalweight concrete slabs. This system is now little used partly owing to excess weight ofthe concrete and partly owing to the slow construction time and the need for proppingthe deck often over two storeys. The problems with propping can be reduced by the useof proprietary falsework systems involving lightweight trusses which allow the actualformwork to be struck before the props are removed. The use of in-situ concrete floorsystems except for small areas not otherwise able to be handled has been supersededby pre-cast pre-stressed concrete units or by composite steel–concrete decking.

Pre-cast pre-stressed concrete units may be placed on the top flange of the steel beam,supported on shelf angles welded to the web of the beam, or placed between theflanges of a column section used as a beam. In all cases the flange of the supportingbeam system may be considered to be restrained against lateral torsional buckling,although for shelf angle floors consideration may need to be given to the torsionalload on the beam during construction. An alternative for ease of construction is touse asymmetric sections with the bottom flange larger than the top (Mullett, 1992;Lawson et al., 1997). These alternatives are illustrated in Fig. 8.1.

For pre-cast units placed on the top flange small discrete vertical plates are welded tothe top flange to give shear anchorage. For floor systems where the pre-cast unit sitson a bottom flange or shelf angle, the top corners of the units will need chamferingto ensure the units can be placed in position. Such floors also have the advantage ofincreased fire performance as the top flange is shielded by the concrete section fromthe effects of a fire on the underside. Equally the bottom flange of the steel sectionwill be either partially or totally adjacent to a heat sink formed by the concrete slab,thus also inducing lower temperatures than would otherwise exist.

For most types of flooring a screed is required to give a proper finish and it is recom-mended that a light structural mesh reinforcement is placed over the support beams tohelp control cracking in the screed. One potential disadvantage of using pre-cast unitsis the large amount of hook time required to place the units which could otherwisehave been used to hoist materials required for finishes, etc.

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Pre-cast units

Shelf angles

(a) Shelf angle floors (screed omitted)

(b) Pre-cast units resting on bottom flange of column section (screed omitted)

(c) Asymmetric section with deep trough decking

in-situ concrete

Deep trough decking

FIGURE 8.1 Types of Flooring

Composite steel–concrete floors use thin gauge trough steel sheeting which initiallyacts as permanent shuttering to the concrete before acting compositely after theconcrete has set to provide both tensile and shear ‘reinforcement’ to the finished con-crete slab. The concrete used is lightweight concrete with a specific weight of around18–20 kN/m3, and designed to be placed using concrete pumps rather than skips. Theonly cranage required is that to lift the bundled steel sheets to the correct level, asthe sheets are individually light enough to be manhandled. To provide shear conti-nuity between the deck and the steel support beams in-situ through deck shear studwelding is employed. Reinforcement will be required in the form of mesh in the topface of the slab in areas of hogging moments if only to control the effects of crack-ing. This mesh will also contribute to the fire performance of the deck. Any proppingcan be avoided by limiting the spans of the decking whilst acting non-compositely byadditional support beams. This option is more cost effective than propping. A recentdevelopment is to use deep decking over the whole depth of a UC used as a beamwith the decking supported on flange plates welded to the lower flange of the UC orspecially rolled asymmetric sections. This system is known as ‘Slimflor’ construction(Mullett and Lawson, 1993), in which the outer beams are downstand UB’s or rolledhollow sections with plates welded to the soffit (Mullett, 1997). A very useful practical

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guide for an overview of design and construction is published by the SCI (Couchmanet al., 2000).

8.3 INFLUENCE OF CONNECTION DESIGN AND DETAILING

From the previous chapter it will be noticed that connections are either designed totake the effect of beam reactions in the form of shear (with nominal moments) or toresist the effects of moments, and the coincident shear and axial force. The first typeis generally designated a ‘pin’ joint which allows large relative rotations between themembers in the connection (and leads to the concept of simple construction). Thesecond is generally designated a ‘fixed’ or ‘rigid’ joint in which rotational compatibilityexists between the members framing into the connection. This type of connectionallows full moment transfer and should ideally be welded, although in UK practice aheavily bolted stiff connection is taken as rigid. Obviously the type of connection in thestructure will markedly influence the behaviour of the structure under whatever actionsare applied to it. A fuller discussion of the implications of the above paragraph is givenin Section 8.12, but it is first necessary to consider the actions applied to a structure.

8.4 STRUCTURAL ACTIONS

These may either be physical loading or imposed deformations due, for example, todifferential settlement.

8.4.1 Physical Loading

For convenience this is divided into two categories: gravity and non-gravity loading.

8.4.1.1 Gravity Loading

This covers the self-weight of the structure, the finishes on the structure, the actions dueto the usage of the structure (variable actions) and roof loading whether as a nominalload or as snow loading (uniform or drifting as appropriate). Values of loads are given inEN 1991-1-1. It should however be noted that snow drift loading given in EN 1991-1-3constitutes an accidental load case and therefore takes lower partial safety factors, andthus may not be critical. It is advisable to design the structure under uniform roof load-ing and then check the structure, if appropriate, under non-uniform snow drift loads.

8.4.1.2 Non-gravity Loading

This can be considered under a series of sub-headings:

• Wind loading: This is covered by EN 1991-1-4.• Inertia and impact loads: These have to be considered where dynamic loading is

considered, for example cranes and supporting systems (EN 1991-3). Impact loading

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needs considering in the case of car parks where columns can be damaged due tocollisions or for bridge piers where vehicular impact may occur (EN 1991-1-7).

• Seismic loads: These are not of general importance in the UK except for nuclearinstallations and other similar structures, and are covered in EN 1998.

• Accidental loads: These can be due to snow drift loading (EN 1991-1-3), fire(EN 1991-1-2) or explosions (EN 1991-1-7). In the case of explosions either theimplications of progressive collapse needs to be considered or the structure mustbe tied together and the resultant tying forces considered.

8.4.2 Deformations

Structural deformations can be either due to differential settlement (Section 8.11.1)or thermal movements (EN 1991-1-5) (Section 8.11.2).

8.4.3 Load Combinations

Load combinations for single variable loads have been covered in Chapter 3, and ittherefore remains to cover load combinations for multiple variable loads. The classicexample of this is where there are variable loads due to structure usage (e.g. officeloading) and variable loads due to wind. The basis behind the combination rules in EN1990 is that it is deemed statistically unlikely that all the variable loads will be acting attheir maximum intensity at the same time. It should be noted that roof loading and floorloading taken for the structure as a whole are considered as separate variable actions.The rules take account of the non-simultaneity of maximum effects by introducing ψ

factors on the non-principal variable actions.

EN 1990 allows a number of combination rules – this is reflected by the UK NationalAnnexe to EN 1990.

At ultimate limit state, the following combination rules are available:

1,35Gk + 1,5Qk,1 +n∑2

1,5ψ0, jQk, j (8.1)

where Gk is the permanent load, Qk,1 is the principal (or leading) variable load and Qk,2

to Qk,n are the accompanying variable actions. Note however that where the permanentload is unfavourable (i.e. is of opposite sign to the permanent load, the partial safetyfactor γG is set equal to 1,0, and where the variable loading is unfavourable its partialsafety factor γQ is taken as zero. Equation (8.1) must be applied taking each variableload as the principal variable load in turn and the remainder as accompanying actions.Experience, however, may be used in reducing the calculations when it is clear whichleading action is critical. Note it is possible for different types of actions to havedifferent values of ψ0, j.

1,35Gk + 1,5ψ0,1Qk,1 +n∑2

1,5ψ0, jQk, j (8.2)

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It is necessary with this combination to examine the effects of different ψ0, j values,although it is likely to give lower loads than those determined from Eq. (8.1).

1,35ξGk + 1,5Qk,1 +n∑2

1,5ψ0, jQk, j (8.3)

The additional factor on the permanent load ξ is subject to the limit 0.85 ≤ ξ ≤ 1.0. TheUK National Annexe specifies a value of 0,925. Equation (8.3) will give the lowest totalload provided the permanent load is unfavourable. It would appear permissible to usethe ξ factor where there is only a single variable load.

For serviceability loading the combination rules are similar except the maximum valuesof γG and γQ are set equal to one as is ξ. The other change is that ψ0 may be replacedby ψ1 or ψ2. The factor ψ1 is used to determine effects under reversible limits states(known as the frequent combination) and ψ2 is used to determine long-term effectsor the appearance of the structure (known as quasi-permanent). The use of ψ0 is onlyrequired for irreversible limit states.

Deflection is generally checked under the quasi-permanent combination although itwould be advisable to check it under full variable load together with permanent loadin order to assess any maximum instantaneous deflection.

It is now necessary to consider how structures resist the forces due to actions appliedto the structure. The major concern is with the transfer of horizontally applied forces,although some consideration is given to the distribution to supporting members ofvertically applied forces. Single storey structures will be considered in detail, beforecontinuing by looking at multi-storey structures.

8.5 SINGLE STOREY STRUCTURES UNDER HORIZONTAL LOADING

Consider initially a basic structure comprising a single bay flat roof portal type framewith encastré feet under vertical loading only (Fig. 8.2). It does not at first sight matterwhether the connections at B and C are rigid or pinned, as the structure can be analysedand the members designed under the resultant forces. However if the joints at B and Care pinned the rafter will be a heavier section as no ‘fixing’ moment will exist at theconnections, and there will be an increased deflection at the centre of the rafter. Thecolumn section will be lighter and the connection detail simplified.

However, the bottom ends of the column are fixed with respect to the bases. Thiscondition will cause problems for the foundation design (Section 8.11.1) as the ratioof moment to axial load will be too high to give an economic design. The generalcustom on such a frame is therefore to use non-moment resisting connections betweenthe feet of the stanchion and foundations, thus if non-moment resisting connectionsare used between the rafter and the columns the frame is inherently unstable. This

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Rigid joint

Vertical loading

A D

CB

(a) Basic frame with rigid joints at B and C

Pins

A D

CB

(b) Basic frame with pinned joints at B and C FIGURE 8.2 Simple single bay flat roofportal

statement is clearly true for horizontal loading, but is also true for vertical loadingdue to inherent imperfections in both the members and through construction. Twosolutions, not considering moment resistant connections at the feet of the columns,are possible. The first is to restore the moment connections at B and C; the second isto find an alternative method to resist any horizontal forces.

The first method has the drawback that for single bay frames the total moments atthe connections may reverse in sign due to the change in direction of the applicationof wind forces. The moment due to the wind force alone must change sign (Fig. 8.3).Allowing the connections to resist wind forces is possible in multi-bay structures wherethe wind force is adsorbed through a large number of connections, thus reducing themoment on each individual connection (see Section 8.12).

The second method is to adopt a bracing system within the frame. So for the windforce P (Fig. 8.4(a)), a diagonal member is placed such that the diagonal member is intension (i.e. capable of taking full design load). This has the effect of triangulating thestructure. When the wind blows from the reverse direction (force P′ in Fig. 8.4(b)),the tie AC now becomes a strut with much reduced load capacity owing to buckling.Thus a further tie BD is now inserted giving rise to cross-bracing. The cross-bracingis designed by only considering the forces in tension members (i.e. the compressionmembers are assumed to have zero load capacity, a conservative assumption). It shouldalso be noted that since the structure is triangulated the frame analysis (and design)is much simplified.

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(a) Wind from left to right

(b) Wind from right to left

A D

A D

B C

C

B

P(Wind force)

P ′(Wind force)

ResultantBMD

FIGURE 8.3 Effect of windreversal

P(Wind force)

P ′(Wind force)

Tie to take wind force P

Tie to take wind force P ′

B

A D

A D

C

B C

(a) Triangulation to take effect of wind force P

(b) Triangulation to take effect of wind force P ′

Note: AC now acts as a strut, and for calculating the force in BD is assumed to have zero strength

FIGURE 8.4 Elementary wind bracing

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(a) Single storey multi-bay frame – each frame braced

(b) Single storey multi-bay frame with wind girder

Eaves beam

FIGURE 8.5 Wind girder bracing

No structure exists as a single frame. Where a set of such frames described above, com-plete with cross-bracing, to be assembled as multi-bay single span structure, it wouldbe unusable (Fig. 8.5(a)). This state of affairs can be made acceptable by retainingthe cross-bracing in each of the end frames only and by supplying full diagonal cross-bracing between each of the frames at rafter level (Fig. 8.5(b)). Such a bracing systemis known as a wind girder, and transmits any horizontal forces through the girder to

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Screed

Pre-cast units

Beam

(a) Section

(b) Elevation (pre-cast units omitted)

Intermittentsteel flatswelded totop flange

FIGURE 8.6 Shear key detail

the bracing in the end frames. A wind girder is always needed where lightweight roof-ing systems are used in conjunction with light gauge purlin systems. Where stressedskin construction is used the wind girder can be omitted as the sheeting is designed toresist the shear due to the horizontal loadings (Davies and Bryan, 1982). If pre-castconcrete units are used then these may replace the wind girder provided adequateshear connection between the rafters and the units is provided (Fig. 8.6). Compositedecking with through deck shear studs will not generally need additional bracing (butsee Section 8.9.3).

For multi-bay multi-span structures a wind girder is needed around the periphery ofthe structure and full diagonal bracing in each of the corners of the structure (Fig. 8.7).For portal frame systems, the wind forces are taken by moment resisting connectionsin the plane of the frame, but for wind applied along the structure, i.e., normal to theplane of the portals, wind bracing must be provided. It may also be necessary to bracethe end or gable frame for wind in the plane of the frame if it is designed as raftersspanning over intermediate gable end columns. Although this section has referred todiscrete bracing it is possible to replace the bracing by alternatives. The most commonreplacement for vertical bracing is masonry shear wall construction. For single storeyconstruction this need be no more than conventional masonry cladding provided suchcladding is fully ties to the steel frame using, for example, half wall ties spot welded tothe stanchions or approved proprietary tying systems. The design of such cladding is toEN 1996-1-1. It should however be remembered that in the case of masonry cladding

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Key :Main members;Wind bracing

Note : All corners must be fully braced

FIGURE 8.7 Schematic layout of wind bracing for a multi-bay structure

the frame is erected first and that the frame will need temporary bracing during thisstage. It is the designer’s responsibility to ensure this is present (Construction ‘Designand Management’ Regulations) (HMSO, 1994).

8.6 MULTI-STOREY CONSTRUCTION

The situation with respect to vertical and horizontal planes can be discussed separately.

8.6.1 Vertical Plane

The effect of horizontal forces in the vertical plane can be handled in a number of ways:

• BracingThe structure can be braced either with the bracing left as an external feature andhidden behind lightweight cladding. Such bracing can be at each corner of the struct-ure, or provided the floors act as stiff diaphragms, in the centre of each face of thestructure. The latter layout is only really suitable for structures symmetric abouttheir centre lines as otherwise torsional effects are introduced which cannot easilybe resisted.

• Shear wallsThis solution is generally adopted for the smaller end panels of structures whichhave a high aspect ratio with few columns in the smaller direction. The shear wallsin order to act properly must be relatively unpierced, i.e., free from dominant or sig-nificant openings. This can produce architectural constraints. Irwin (1984) provides

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information on design of shear walls. In the longitudinal direction the wind force islower (as it acts on the smaller face) and there are a large number of bays to takethe wind forces on the columns and the column–beam connections.

• CoresIn the UK, the most common solution is to allow the wind forces to be adsorbed bythe core(s) provided for the lift shaft(s) and/or stairwell(s). Such cores are generallyof reinforced concrete construction, although the walls surrounding stairwells maybe masonry. The walls surrounding lifts or stairwells do not have large openingsas they generally provide access to fire escape routes and protected access areasfor fire fighting, and thus stiff enough to provide the lateral restraint to the struct-ure necessary. It should be noted that tolerance problems can arise when marryingup steelwork and reinforced concrete construction, and that shrinkage and creepeffects in the concrete should not be ignored (Irwin, 1984).

• Rigid/semi-rigid constructionThe case where the frame is taken as rigid for both the vertical loading and hori-zontal loading is rare in the UK. However, a hybrid method of design is becomingpopular in which the frame is consider pin jointed (simple construction) for the ver-tical loading but as rigid jointed for the horizontal loading. There are restrictions onthis method mostly related to the size and shape of the structure (Salter et al., 1999).

8.6.2 Horizontal Plane

Composite steel–concrete flooring systems will, when the concrete has hardened,prove a very stiff diaphragm to transmit horizontal forces provided adequate shearconnection between the deck and the steel skeleton is available. The shear stud require-ment to provide flexural composite action will generally be adequate. When pre-castconcrete units are used a shear key detail such as that illustrated in Fig. 8.6 shouldbe used, and structural mesh provided in the screed in order to give full diaphragmaction and prevent cracking over the beam.

8.7 BEHAVIOUR UNDER ACCIDENTAL EFFECTS

Accidental actions should be considered in a number of possible circumstances:

• Explosions whether due to gas or terrorism.• Impact due to vehicles or aircraft.

Should the risk of such an incident be high and the effects be catastrophic, or incertain circumstances the need to check be mandatory (e.g. vehicular impact onbridge piers), then the designer must ensure that the structure is designed anddetailed to ensure that should an accidental situation occur, the structure doesnot suffer complete or partial collapse from either the accidental situation itself orsubsequent events, such as spread of fire.

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Four now classic cases where collapse occurred due to accidental or terroristaction are:

(1) Ronan Point block of flats where a gas explosion blew out a wall panel causingprogressive collapse of a corner of the structure (Wearne, 1999).

(2) The Alfred P. Murrah Building in Oklahoma City where collapse was caused by aterrorist bomb blast (Wearne, 1999).

(3) World Trade Centre Towers in New York on 11 September 2001 due to impactfrom deliberate low flying aircraft and subsequent fire spread. The towers weresteel framed structures and it has since been demonstrated that the prime cause ofcollapse was due to loss of fire protection on the floor members, as it was consid-ered that the fire load from the impacting aircraft was not unduly high (Dowling,2005). Information on considerations of the design of high rise construction isgiven in ISE (2002).

(4) Pentagon Building on 11 September 2001 following the deliberate crash of a com-mercial airline. The concrete structure suffered partial but not extensive collapsedue to the impact and subsequent fire (Mlakar et al., 2003).

8.7.1 Progressive Collapse

EN 1990 identifies the need to consider that in the event of an accident such as explo-sion or fire the structure should not exhibit disproportionate damage. This is reinforcedby the requirements of the relevant Building Regulations within the UK, for exam-ple the recently revised Approved Document A of the England and Wales BuildingRegulations.

Such damage can be mitigated by:

• Attempting to reduce or limit the hazardIn the case of fire this could be done by full consideration being given to the useof non-flammable materials within the structure and by the provisional of relevantactive fire protection measures such as sprinklers. In the case of industrial processeswhere explosions are a risk then such potential risks need to be taken into accountby, say, enclosing the process in blast proof enclosures.

• MaintenanceThe recent Pipers Row Car Park collapse of a lift slab concrete structure (Wood,2003) indicated the need for adequate design especially where punching shear in aconcrete slab may be critical as this is a quasi-brittle failure. The need for adequateinspection and maintenance where it is known that environmental effects will besevere was also highlighted. Although the failure was due to poor maintenance, itwas exacerbated by uneven reaction distribution at the slab–column interface andno tying through the lower face of the slab. Admittedly the construction techniqueused would have made the latter extremely difficult although it contributed to theformer.

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• Consideration of the structural formA case where this is relevant is when the structure could be subject to externallyprovoked explosions. One reason why the damage was extensive following the Okla-homa City explosion was that the blast was amplified by an overhanging portion ofthe structure above ground floor level. A solution for structures known to be atpotential risk is to provide an external curtain which is not structural and is eas-ily blown out whilst ensuring the structure is stabilized by a core which is in thecentre of the building. Additionally the risk can be mitigated by ensuring vehiclescannot come within certain limiting distances of the structure. More informationon this, including assessment of blast forces is given in a SCI publication (Yandzioand Gough, 1999).

• Ensuring the structure is adequately tied to resist collapseThis needs considering on two levels. The first is to determine the magnitudes ofthe likely levels of tying force required. The second is to provide properly detailedconnections between elements of the structure. This is needed for both connectionsin the horizontal plane: beam-to beam or beam to column and in the vertical plane:column splices.

• Removal of key elements to establish stabilityIn this approach a risk analysis is carried out to identify the key elements in a struc-ture which are vital for its stability. A structural analysis is carried out with any one ofthese elements removed to establish the stability of the remaining structure and itsability to carry the loading so induced. As this is an accidental limit state lower partialsafety factors are used on both the loading and the material strengths determiningthe member strengths. It is not necessary to check any serviceability limit states.

Approved Document A divides structures into four categories: 1, 2A, 2B and 3, withCategory 1 being the least onerous. Category 1 constitutes low-rise housing, agricul-tural buildings and buildings with restricted access. These require no specific checks.For Category 2A which are medium consequence risk structures, the considerationof tying forces is adequate in most cases. For Category 2B defined as high conse-quence risk structures, then the requirement must be satisfied by using removal of keyelements. For Category 3 structures which are very high consequence risk structuressuch as large capacity grandstands, hospitals, structures over 15 storeys or structuresin which hazardous materials are involved require special consideration (Way, 2005).Alexander (2004) gives examples on risk assessment with the hazards grouped intotwo categories: those that must be considered for any Class 3 structure and those thatonly need considering for specific location dependent structure such as flooding.

8.7.2 Structure Stability

Any structure with high lateral loading or where vertical loading can be applied outsidethe frame envelope must be checked for overturning. Also any continuous memberwith a cantilever must be checked for the possibility of uplift on any support.

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14 m

35 m

1.5 m

gK = 32 kN/mwK = 6 kN/m

WK = 120 kN

A

3 m 6 m GK = 1500 kN

FIGURE 8.8 Design data forExample 8.1

In all these cases the loading which causes overturning is deemed to be unfavourable,and the loading tending to restore the situation is known as favourable. This is definedin EN 1990 as verification of static equilibrium EQU. For this situation the partialsafety factors corresponding to EQU in EN 1990 (Table A1.2 A) must be used.

EXAMPLE 8.1 High wind load.

Consider the possibility of overturning for the water tower shown in Fig. 8.8.

Taking moments about point A:

Unfavourable effects due to wind:

1,5 × 120(3 + 35 + 1,5) + 1,5 × 6 × 35(

352

+ 1,5)

= 13095 kNm

Favourable effects due to self-weight:

0,9 × 1500142

+ 0,9 × 32 × 35142

= 16506 kNm

The moment due to the favourable effects exceeds that due to the unfavourable effect,thus the structure will not overturn.

EXAMPLE 8.2 External loading.

Consider the cantilevering frame shown in Fig. 8.9.

Here the check is more complex, as the permanent load of 220 kN per storey is bothfavourable and unfavourable, in that the load acting on the cantilever is unfavourableas it contributes to the overturning effect. The permanent load per storey may beexpressed as 27,5 kN/m run.

The favourable part of the permanent loading on the internal span takes a factor of0,9, and that on the cantilever of 1,1.

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wk = 13 kN/m 5 @3.5 mGK = 220 kN/storey

B7 m 1 m

FIGURE 8.9 Design data forExample 8.2

Taking moments about B:

Effect due to the wind loading (unfavourable):

1,5 × 1317,52

2= 2986 kNm

Effect due to permanent action (per storey):

0,9 × 27,572

2− 1,1 × 27,5

12

2= 591 kNm

Total restoring (or favourable) moment:

5 × 591 = 2955 kNm

The structure will just overturn as the disturbing moment is marginally greater thanthe restoring moment. Thus either the structure needs tying down using, say, tensionpiles or ground anchors, or the permanent loading could be marginally increased. Thelatter is the cheaper option owing to the small margin involved.

8.8 TRANSMISSION OF LOADING

8.8.1 Transmission of Loading from Flooring Systems

When the loading is considered as a UDL, then the distribution of such loading to anysupporting beam system will depend whether the decking system may be consideredas one or two way spanning. Deck systems comprising lightweight proprietary systems(roofing only), pre-cast concrete slabs, composite steel–concrete decks and timber areall one way spanning. Thus half the load is taken to either end of the pre-cast unit,purlin system, joists or composite deck (in the direction of the troughs) (Fig. 8.10).Where a decking system is continuous over the supporting beam system then the reac-tions should be determined using a suitable analysis, although for approximate designcontinuity may be ignored. Where in-situ concrete is used then loading is distributed to

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Reaction from decking (UDL)

Main beams

Reactionfrom joistReaction

from joist

Decking

Joist orpurlin

Note : The loading on the main beam will be appiled through web cleats (not shown)

Reaction (UDL)

Concrete slab withprofiled sheet steel decking

Mainbeam

Intermediatebeam reactions

Load onmainbeam

(a) Transfer of load from one way spanning concrete deck

(b) Load transfer from decking through a joist or purlin system

FIGURE 8.10 Load transfer for one way spanning system

45°

D C

A

E F

B

Note : (a) Load on area AED is supported by beam AD (b) Load on area AEFB is supported by beam AB

FIGURE 8.11 Approximate loadtransfer from two way spanningslabs

all the supporting beams. The loads on individual beams should be determined usingthe same assumptions as in the slab design, such as Johansen’s yield line method orthe Hillerborg strip approach. However, for approximate calculations a 45◦ disper-sion from each corner of the slab may be used to partition the loading to the beams(Fig. 8.11).

8.8.2 Lintels

In the absence of other data a 60◦ dispersion of the load above the lintel may be used,that is, the lintel only carries the loading within the complete 60◦ equilateral triangle.Where the triangle is incomplete due to openings then the whole load above shouldbe taken.

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8.9 DESIGN OF BRACING

8.9.1 Permanent Bracing

The forces to be used to design such bracing are given in cl 5.3:

(a) any horizontal loads applied to the frame being braced,(b) any horizontal or vertical loads applied directly to the bracing system(c) the effects of initial imperfections, or the equivalent horizontal forces, from

the bracing system itself and from all the frames it braces.

For wind bracing the forces in (a) and (b) will need considering, whereas for swaybracing in the vertical plane it is a combination of all three. The equivalent horizontalforces for sway are covered in Section 8.12.2. There is an additional requirementfor stability forces at the splice in a column or beam which is covered in the nextsection.

The design of diagonal bracing is generally simplified by ignoring diagonal membersin compression and assuming all joints are pinned. In the case of wind bracing themember should not be slender enough to induce an acceptable degree of sag underits own self-weight.

8.9.2 Restraint Bracing to Compression Flanges andColumn Splices

• Compression flange bracingThe problem can either be handled by adding an additional bow in the mem-bers to be restrained and designing for the resultant additional moments, or usingan equivalent stabilizing force. In the first case the initial bow imperfection e0 isgiven by

e0 = αmL

500(8.4)

where L is the span of the bracing system and αm is given by

αm =√

0,5(

1 + 1m

)(8.5)

where m is the number of members to be restrained.For convenience the effective bow imperfection in the members to be restrained bya bracing system may be replaced by the equivalent stabilizing force qd is given by

qd =∑

NEd8e0 + δq

L2 (8.6)

where δq is the in-plane deflection due to the load q and any external loads calculatedfrom a first order analysis. If second order analysis is used then δq = 0.

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The notional force NEd in the flange is defined as MEd/h where MEd is the maximummoment in the flange and h is the overall depth of the beam.

• Restraint to splicesAn additional requirement at splices is that the bracing should be able to resista local force of αmNEd/100 together with any other applied forces. Note that theapplication of this local force is not co-existent with the force q defined above.

8.9.3 Temporary or Erection Bracing

Under the Temporary Workplaces Directive enforced in the UK by the Construction(Design and Management) Regulations, the client must appoint a planning supervisorat the Design stage on all health and safety matters throughout the execution of theproject. It thus falls to the supervisor to ensure the structure is capable of being erectedsafely and that the requisite safety measures are in force. This means that the speci-fication and design of temporary bracing should not be left to the steelwork erectorwithout approval by the planning supervisor. The necessity for such bracing is oftendue to the intended construction sequence where construction starts at the centre ofa structure in order to reduce cumulative tolerance errors whereas the permanentbracing designed to give stability to the completed structure is in the end bays. Furtherreasons are that support conditions or the effect of temporary loadings may be suchthat tension members are in compression or the signs of bending moments are reduced.There is also the requirement to ensure that any forces applied to the structure fromsafety equipment used during erection are considered. Where in-situ concrete is beingused the compression flange will not receive full restraint until the concrete is hard-ened, and thus during construction the compression flange of the support beam systemis unrestrained and therefore prone to lateral torsional buckling. Where profiled sheetsteel decking is being used provided through deck welded shear studs are used, thebeams running normal to the deck profiles can be considered as fully restrained, butthose running parallel with the profiles will need checking for buckling though thedeck will provide some degree of restraint (Lawson and Nethercot, 1985). An addi-tional point to be noted is that large members may not be stable when being lifted intoposition partly due to the change in support conditions and partly due to the changesin restraints. Such members may need to be braced together at the correct spacing andlifted in pairs. Such bracing should if possible form part of the permanent bracing.

8.10 FIRE PERFORMANCE

It is recommended that the reader is referred to Ham et al. (1999) for an overviewof the basic procedures and of methods of protection. Whilst it is not intended tocover detail fire design of steel structures, a brief overview is necessary. For full detailsreference should be made, for example, to Purkiss (2007).

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8.10.1 Single Storey Structures

For most single storey structures the situation with respect to fire performance is rela-tively straightforward as there is generally no specific requirement within the Englandand Wales Building Regulations (Approved Document B) (DTLR, 2000). The onlyproblem tends to be caused by outward collapse of walls and columns of single storeyframes. This tends to be worse with pitched roof portals (see Simms and Newman,2002). In addition there is the problem of ensuring that fire fighters have completesafe access to fight the fire during its complete duration.

8.10.2 Multi-storey Structures

Traditionally design has been by considering single elements with no interactionbetween such elements. However, fire tests on the steel frame structure at Cardingtonhave demonstrated that with structural interaction between beams and columns it ispossible to leave all beams unprotected (except at connections) and achieve a verygood fire performance. Interim design data allowing some beams within floor systemsbeing left unprotected has been published by the SCI (Newman et al., 2006). It isalso possible to encase columns in brickwork or blockwork or to fill rolled hollow sec-tion columns with concrete to achieve fire performance without additional measures(Bailey et al., 1999).

Although a complete discussion of the Cardington tests is given elsewhere (Purkiss,2007), an overview of the tests will be presented. The tests at Cardington demonstratedthat temperatures in the unprotected steel beams which were composite with the floorslab reached temperatures on the bottom flange in excess of 1000◦C. This is around400◦C higher than the limiting temperatures allowed after correcting for load level indesign codes. It should be noted that the loading applied on the floors was around one-third of the ambient design value of 2,5 kPa giving a load ratio lower than that of mostoffice type structures. Although the structure retained its load carrying capacity thedeflections were extremely high with values of up to 640 mm being recorded. Althoughthe tests showed far better performance than would be expected from the standardfurnace test on individual elements, a number of points need to be raised:

(1) The high deflections reached would probably mean part at least of the structurewould need replacing.

(2) Early tests in the series were performed with unprotected columns which sufferedsevere buckling just below the beam columns connections.

(3) Deflections of some beams were more than sufficient to cause internal lightweightcompartment walls to fail leading to loss of compartmentation and thus increasedfire spread (Bailey, 2004).

(4) Buckling in the lower flanges of the beams occurred at the ends owing to high-induced compressive forces as the moments were redistributed away from mid-span. This has led to the recommendation that design allowing for increased

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moment capacity at the connection (Lawson, 1990a, b) should not be used wherethe beam is restrained against lateral movement during a fire (Bailey et al., 1999).

(5) During the cooling phase the beams do not recover plastic deformations andpotentially thus place the connections in tension. This in fact resulted in the partialfailure of some connections due to shear in the bolts, excessive flexure of end platesor failure in the welds connecting the end plates to the beams (Bailey, 2004).

Design methods have, however, been introduced which allow for some secondarybeams in composite construction to be left unprotected (Newman et al., 2006) andwhich allow a better analysis of composite floor slab action under the compressivemembrane force system induced in the fire limit state (Bailey (2001, 2003)).

Even with the traditional approach shelf angle beams will give 30 min standard fireperformance and may give 60 min (Newman, 1993). Slim floor construction should give60 min with no additional fire protection. Under no circumstances should columns beleft unprotected. For fire performance periods of 30 min infilling the web space of aUC with non-structural blockwork can be sufficient (Newman, 1992).

8.11 ADDITIONAL DESIGN CONSTRAINTS

8.11.1 Ground Conditions and Foundations

This section is intended to be neither a comprehensive survey of foundation design orconstruction nor of problems associated with ground conditions. It is offered as an aide-memoir and covers the implications of any problematic areas. For a full consideration,the reader is referred to Henry (1986). Also the design of concrete foundations iscovered in Martin and Purkiss (2006).

The selection of foundation type generally depends on the potential bearing capacityof the ground and its susceptibility to absolute or differential settlement. The mosteconomic form of foundation is the simple pad foundation which in order to be of areasonable size requires a reasonably high bearing capacity. If bearing capacities arelow, and the loads applied to the foundations are high, the individual pad foundationswill start to overlap and will become either combined foundations or in the extremecase raft foundations. Where the bearing capacity is low then either piled foundations,ground stabilization, or the use of replacement imported fill should be considered. Thedecision will be primarily based on cost, although environmental considerations suchas noise, dust and extra traffic must be taken into account.

Ideally pad or raft foundations should be designed for uniform bearing pressure atultimate limit state under imposed loading other than wind or accidental actions. Thiswill generally be possible for raft foundations, but may not be for pad foundations.It is recommended that for pad foundations the ratio of the maximum to minimumbearing pressures should not exceed around 1,5. In order to achieve this it is possible tooffset any stanchion away from the centre line of the foundation except where severe

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moment reversals may occur. It also leads to the adoption of nominally pinned feetwhere the axial load in single storey structures is low as the required ratio cannot besatisfied with fixed feet. Uplift should never be allowed below a foundation undernormal conditions. It should be noted that it may be necessary, in practice, to supplya degree of fixity to bases of columns in single bay portal frames when consideringsuch frames at site boundaries in the case of fire and thereby to accept a high ratio ofmaximum to minimum bearing pressures, even uplift, as this is an accidental situation.

Any horizontal reaction at the base of a column will also apply a moment to the under-side of the foundation and must be included. If the structure is tied together by groundbeams then only the net horizontal force need be considered, otherwise each reactionmust be considered independently. For multi-storey structures it is general practice toemploy tower cranes to facilitate erection. These need substantial foundations whichare often incorporated into the structure as the foundations to the lift shafts.

Differential settlement may either be caused by non-uniform loading imposed by thestructure on its foundations or by variable ground conditions beneath the foundations.The effect of non-uniform loading can be mitigated by placing the whole structureon a single raft foundation, and thus producing sensibly uniform bearing pressures.Also to avoid differential movement between any masonry cladding and the frame,the masonry can be supported on ground beams tied to the pad foundations belowthe columns. Alternatively the components of the structure can be allowed to settleindependently by the provision of movement joints, in which case the cladding maynot be considered to act as bracing and full wind bracing will need to be designed.

Where the structure is such that differential settlement cannot be avoided then eitherthe structure must be designed to accept these movements (note, differential settle-ment does not affect moments when determined using plastic analysis; they do whenan elastic analysis is used), or the structure can be articulated such that the movementsdo not affect the internal forces by the provision of hinges, often making the structureiso-static. Typically such hinges should be at the points of contraflexure, and must bechecked for rotational capacity expected through such movements (e.g. Fig. 8.12). Thistype of articulation is frequently used on bridge decks where there may be substantialdifferential settlement due often to mining subsidence. Often in this case where suchsubsidence could be large, provision is built in to jack up the structure, in which casea realistic estimate will be needed of such movements.

8.11.2 Expansion and Contraction Joints

It can be a matter of some debate on whether these should be provided and if so atwhat spacing. It is suggested that masonry should have movement joints at around 7 mcentres (or the frame spacing), and concrete floor slabs at around 20–30 m (Deacon,1986; Bussell and Cather, 1995; Concrete Society, 2003). This suggests that continuousmulti-bay frames should have movement joints at around the same spacing. Thus either

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UDL

Note : BMD is drawn on the tension face on the beam

Hinges or flexible joints at positionsof points of contraflexure

BMD

FIGURE 8.12 Cantilever-suspended span structure

movement joints should be provided through the whole height of the structure, in whichcase full wind bracing needs providing in each portion of the structure, or the structurewill need designing to take the actions caused by such movements. For informationon movement joints see Alexander and Lawson (1981).

8.11.3 Stability

Stability can be considered at two levels: member stability and frame stability.

• Member stabilityThis effectively ensures that there will be sufficient ductility in the member to ensurethat plastic rotational capacity is available when required by the design/analysissynthesis. The ductility check is made by considering limiting values of flange andweb slenderness. It is essential that where plastic analysis is used Class 1 sections aremandatory, although where only a single hinge is needed for collapse, as in a simplysupported beam, Class 2 sections may be used. It should be noted that virtually allUKBs are Class 1 whether Grade S255 or S335.The assumption is made in plastic analysis that the member can achieve its fullplastic moment capacity before the onset of elasto-plastic buckling. For members insimple construction or isolated beam elements, this condition is not necessary as theload carrying capacity of the member can be checked using reduced strengths whichallow for buckling. For rigid jointed frames premature elasto-plastic buckling is notpermissible and must be counteracted either by bracing which reduces effective, orsystem, lengths below critical values or by increasing member sizes.

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• Frame stabilityFor portal frame systems this takes the form of two checks. The first is the determi-nation of vertical deflections at the ridges and horizontal deflections at the top ofthe stanchions at the eaves. The second check is only for multi-span portals and isdesigned to avoid ‘snap through’ of the rafters.For multi-storey structures both the relative lateral deflections on each storey andthe overall lateral deflection of the whole structure is subject to limits.

8.12 DESIGN PHILOSOPHIES

The basic analysis methods allowed are elastic or plastic (the use of elasto-plastic meth-ods may be needed where sway is important). Traditionally the UK has adopted simpleframing where the frame carries the vertical loading and, the bracing system, the hor-izontal loading. Continuous framing (except for portal frames) and semi-continuousframing are rarer.

8.12.1 Frame Classification

A frame may be classified as sway or non-sway. A non-sway frame is one where hori-zontal deformations are limited by the provision of substantial bracing members oftenin the form of lift shafts or stairwells often as part of a central core. A frame withbracing may still be classified as a sway frame.

The criteria for neglecting global second order effects are for

• Elastic analysis

αcr = Fcr

FEd≥ 10,0 (8.7)

• Plastic analysis

αcr = Fcr

FEd≥ 15,0 (8.8)

where αcr is the ratio by which the design loading would have to be increased to causeglobal elastic instability, FEd is the design load on the structure and Fcr is the elasticcritical global buckling load. The higher limit for plastic analysis is due to the issuethat imperfections become more important in plastic analysis due to P– δ effects.

For shallow pitch portal frames (roof slope less than 26◦) or beam and column typestructures αcr is given by

αcr = HEd

VEd

hδh,Ed

(8.9)

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where HEd is the total horizontal reaction at the bottom of a storey (including notionalloads), VEd is total vertical loading at the bottom of a storey, δh,Ed is the horizontaldisplacement due to horizontal loading including notional horizontal loads and h isthe storey height.

The value of αcr may also be determined by applying nominal geometric rotations,or equivalent horizontal static forces, to the nodes in a frame and determining theresultant horizontal deformations. The disturbing forces are small so that any sway islow and will not cause large deflections. The method is based on the theory proposed byHorne (1975) for the determination of the elastic critical load factor. A braced framecannot a priori be considered as a non-sway frame. Determination of sway classificationis either carried out by imposing a rotation of ϕ′ at the foot of each stanchion or byimposing horizontal forces of ϕ′W at each level of the frame where W is the totalfactored vertical loading at that level. For each level the sawy index �/h is determined.From Horne and Morris (1981) the value of αcr is given by the minimum value of

αcr = 0,009(�h

)max

(8.10)

For multi-storey frames with diagonal bracing Zalka (1999) provides a very quickway of estimating αcr. The critical frame buckling load Ncr for a frame loaded withuniformly distributed loading is given by

Ncr = λK (8.11)

where the parameter λ is tabulated in terms of a parameter βs defined by

βs = KNg

(8.12)

where Ng is the global bending critical load given by

Ng = nn + 1,588

7,837EcIg

H2 (8.13)

Where n is the number of storeys, H is the height of the building and EcIg is the globalflexural rigidity of the columns determined using the parallel axis theorem with thesecond moment of the columns themselves being neglected.

The shear stiffness K is dependent upon the type of bracing.

For single direction bracing,

K =(

d3

AdEdhl2− 1

AhEhh

)−1

(8.14)

and for complete diagonal cross-bracing,

K = AdEdhl2

d3 (8.15)

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00

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

2 4 6 8 10 12 14 16 18 20

βs

λ

FIGURE 8.13 Critical load parameter λ (Zalka, 1999) by permission

where AdEd is the axial rigidity of the bracing member, AhEh is the axial rigidity ofthe horizontal member, d is the length of the diagonal member, l is the width of thebraced bay and h is the storey height.

The formulations of K for other bracing layouts are given in Zalka (1999). Figure 8.13from Zalka (1999) gives the relationship between βs and λ (the data are also given intabular form in Zalka).

8.12.2 Frame Imperfections

For frames sensitive to sway buckling, an allowance in the frame analysis should bemade by introducing and initial sway imperfection and an initial bow.

8.12.2.1 Initial Sway Imperfection

The imperfection factor φ is determined from

φ = αhαmφ0 (8.16)

where the reduction factor for the building height αh is given by

αh = 2√h

(8.17)

where h is the height of the building (in m) and 2/3 ≤ αh ≤ 1,0.

The factor αm depends on the number of columns m in a row,

αm =√

0,5(

1 + 1m

)(8.18)

The basic imperfection factor φ0 is equal to 1/200.

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Note, the effects of this may be ignored if HEd ≥ 0,15VEd, provided the normalizedslenderness ratio λ complies with

λ ≥ 0,3

√AfyNEd

(8.19)

8.12.2.2 Initial Bow

This is expressed as the initial bow e0 divided by the length l. It is a function of thestrut buckling curve and whether elastic or plastic methods of analysis are used. Thevalues are given in Table 5.1 of EN 1993-1-1. These effects should however be ignoredif the members are checked for buckling.

8.12.3 Simple Framing

In simple framing the joints between the members may be assumed to offer negligiblemoment transfer, and the beams should be designed as simply supported. This meansany horizontal loading must be taken by bracing or any equivalent substitute. Simpleframing can be used to design conventional multi-storey beam and column structuresor triangulated truss systems where there will be no significant secondary momentsdue to joint fixity.

8.12.4 Continuous Framing

In this case the connections are designed to give full rigidity between members at a joint.Note on sway and non-sway frames. Elastic, elasto-plastic or full rigid plastic analysesmay be used. It is suggested that reference should be made to Ghali and Neville (1989),Coates et al. (1988), Horne and Morris (1981), Neale (1977) or Moy (1996).

8.12.5 Semi-continuous Framing

There are two basic approaches to semi-continuous framing: the first is to employmoment capacity/rotation characteristics and the second is a hybrid between simpleand continuous.

The basic approach is to use an analysis which incorporates the connection charac-teristics in the analysis (Roberts, 1981; Taylor, 1981). In an elastic analysis the actualmoment–rotation characteristic is input as an equivalent spring at the joint, in a plasticanalysis the moment capacity is used as a local plastic moment capacity, provided theconnection has sufficient ductility.

In the hybrid method, the frame is designed as simple for the purpose of determiningthe member sizes required to carry the imposed vertical loading, but as continuouswhen determining the effects of lateral, or horizontal, loading on the structure (Salter

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et al., 1999). There are limitations on the method based on number of storeys andlayout, but it can be clearly advantageous as it will avoid the need to provide anywind bracing, but at the expense of providing connections capable of transmitting themoments due to wind.

An alternative approach is to use an elastic analysis and then redistribute the momentsto simulate plastic action. EN 1993-1-1, cl 5.4 4(B) places restrictions by specifying thatthe maximum redistribution is limited to 15%, the section classifications of memberswhere moments are reduced must be no worse than Class 2, and that lateral torsionalbuckling must be prevented.

8.13 DESIGN ISSUES FOR MULTI-STOREY STRUCTURES

Brown et al. (2004) provide a useful overview of constructional details for multi-storeyframes. A distinction needs to be drawn between frames which are rigid jointed underall applied loading and frames in which the connections are either semi-rigid or pinned.With pinned connections it should be reiterated that the frame requires bracing toresist the effects of any horizontal actions on the frame, and that even if a frame isbraced it cannot be assumed that the frame is a no-sway frame (Section 8.12).

8.13.1 Simple Construction Frames

In simple construction frames, the beams are designed as simply supported and thecolumns designed to take the beam reactions together with nominal fixing moments.The nominal moments should be calculated assuming the beam reactions are appliedat a minimum of 100 mm from the face of the column for major axis loading. For minoraxis loading, which should be much smaller, the load is applied at either 100 mm fromthe web or the edges of the flanges depending upon the connection detail. If thereaction at the connection acts at a fixed point then this distance should be taken ifit exceeds 100 mm. The nominal fixing moments may be distributed to the lengths ofthe columns immediately above and below the level being designed. There is no carryover to other storeys outside those above and below. The distribution should be inproportion to the stiffness of the column segments.

8.13.2 Semi-rigid Frames

Two methods are here available, either the frame is analysed using actual or design con-nection moment–rotation characteristics and the frame designed in the same manneras a rigid frame, or a hybrid method is used combining simple and continuous con-struction. The latter method has already been briefly outlined (Section 8.11.3), butfurther comment is necessary. The method has the effect of reducing the beam cross-section and the beam deflection. This will ensure that the combined effects of wind,permanent and variable loads will not be critical on the beam itself. The columns need

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to be designed to resist the worst effects of either the permanent and variable loadsor the combination of wind, permanent and variable. In both cases the nominal fix-ing moments from the beams together with the moments induced by the nominalframe imperfection loads must be considered, and may well produce a heavier columncross-section than for simple construction.

8.13.3 Continuous Construction

For continuous construction either elastic or plastic methods may be used to carry outthe analysis. There is a restriction that the compression flange at the hinge must beClass 1, and where a transverse force exceeding 10% of the shear resistance is appliedto the web at the position of the plastic hinge then web stiffeners need to be suppliedat a distance of h/2 from the hinge where h is the beam depth (cl 5.6 2b). For varyingcross-section members, the web thickness should not be reduced for a distance 2deither side of the hinge (where d is the depth between fillets), the flanges must beClass 1 for the greater distance of 2d or the point where the moment drops to 80% ofthe plastic moment at the hinge.

Restraints must be provided at rotated plastic hinges (i.e. all but the last one to form)(cl. 6.3.5.2). This may be provided by diagonal bracing from the lower flange to beamsor purlins fixed to the top flange. Where members carry moment only, or momentwith axial tension, and are connected to a concrete slab by shear connectors, then thisis deemed sufficient for rolled ‘I’ or ‘H’ sections.

At each hinge the restraint and connections should be capable of resisting a force equalto 0,0025Nf,Ed where Nf,Ed is the axial force in the compression flange. In addition tothe imperfection force given by Eq. (8.6), the bracing system should be able to carry aforce Qm is given by

Qm = 1,5αmNf,Ed

100(8.20)

The lateral torsional buckling check of segments between restraints need not be carriedout if the length between restraints is less than Lstable and the moment gradient is linear,where Lstable is given as

for 0,625 ≤ ψ ≤ 1,0

Lstable = 35iz

√235fy

(8.21)

and for −1,0 ≤ ψ ≤ 0,625

Lstable = (60 − 40ψ)iz

√235fy

(8.22)

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where the moment ratio ψ is given by

ψ = MEd,mim

Mpl,Rd(8.23)

Note that where a rotated plastic hinge location occurs immediately adjacent to theend of a haunch, the tapered section need not be checked for stability if the restraintis placed at h/2 along the tapered section (not the uniform section) and the haunchremains elastic along its length.

8.13.4 Effective Length Factors for Continuous Construction

For either elastic or plastic analyses, the design of the columns takes account of therelative joint stiffnesses to determine the effective length factor for the column beingdesigned. The two distribution factors k1 and k2 at the top and bottom of the columnare defined as

k1 = Kc + K1

Kc + K1 + K11 + K12(8.24)

k2 = Kc + K2

Kc + K2 + K21 + K22(8.25)

where Kc is the stiffness of the column, K1 and K2 the stiffnesses of the columns atends 1 and 2, K11 and K12 the stiffnesses of the beams framing in at end 1 and K21 andK22 the stiffnesses of the beams framing in at end 2 (Fig. 8.14).

• Non-sway framesThe available information to calculate the effective length l is given in BS 5950

Part 1: 2000.The ratio of the effective length l to the actual length L is given by either

lL

= 0,5 + 0,14(k1 + k2) + 0,055(k1 + k2)2 (8.26)

Note that the moments due to the nominal frame imperfections must be included inthe design moments for the frame in addition to those caused by any imposed actions.

• Sway framesThe sway mode effective lengths are calculated from

lL

=[

1 − 0,2(k1 + k2) − 0,12k1k2

1 − 0,8(k1 + k2) + 0,6k1k2

]1/2

(8.27)

Williams and Sharp (1990) provide alternative formulations:

lL

= π√12 − 36 k1+k2−k1k2

4−k1k2

(8.28)

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Kc Column being designed

K1

K2

K12

K21 K22

K12

Note : The K values are the stiffnesses of the upper and lower beams, and the upper and lower columns in consistent units.

FIGURE 8.14 Data for determination of column effective lengths

or

lL

= π(1,05 − 0,05k1k2)√12 − 36 k1+k2−k1k2

4−k1k2

(8.29)

Note that for a sway frame l/L may exceed 1.Sway moments should be increased using an amplification factor given by

1/(1 − 1/αcr), provided αcr ≥ 3,0.

8.13.5 Column Loads in Multi-storey Structures

In a multi-storey structure, the frame analysis will provide the reactions in the columns.However, such analyses assume that all floors are fully loaded at all times. Statisticallythis is conservative, thus the axial loads due to variable loading may be reduced in thelower columns of a multi-storey structure. Note, there is clearly no reduction in loadsdue to permanent (or quasi-permanent) loading, and each individual floor must bedesigned under full variable and permanent actions. The variable loading may alsobe adjusted for loaded area, as the loading will be more concentrated over smallerfloor areas (and less concentrated over large areas). This means that column reactionsmay be reduced for this reason also. However, reductions may be made either forthe number of floors or loaded area. These reductions are only for loading categories

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A to D of Table 6.1 of EN 1991-1-1. This prohibits any reduction on situations wherethe loads are predominantly storage or industrial.

The reduction factor αn for the number of stories n (n > 2) is given by

αn = 2 + (n − 2) ψ0

n(8.30)

The reduction factor αA for the loaded area is given by

αA = 57ψ0 + A0

A≤ 1,0 (8.31)

where A is the loaded area, A0 the reference value of 10 m2. The reduction factor isalso subject to the restriction that for load categories C and D it may not be less than 0,6.

8.14 PORTAL FRAME DESIGN

8.14.1 Single Bay Portal

To illustrate the design procedure an internal frame with full rigid connections isconsidered (EXAMPLE 8.3).

• A heavier section will be used for the stanchion than the rafter. Although thisrequires haunches at both the eaves and the ridge to accommodate the connections,the solution is still more economic than the use of a uniform section throughout.

• To avoid stability problems in the eaves haunch the stresses in the eaves are keptelastic by extending the haunch for a distance of span/10 from the face of the stan-chion. The haunch is taken at its maximum depth of twice that of the rafter in orderto ease fabrication.

• Plastic hinges are assumed to occur in the stanchion at the base of the haunch, thatis, 1,5 times the rafter depth below the intersection of the rafter–stanchion centrelines, and at the second purlin point below the ridge. It is then usual to check themoment at the first purlin point.

• The frame will be designed under variable and permanent actions. The frame willnot be checked for wind, as this case is rarely critical on a portal frame.

• Stability will be checked by applying notional forces of ϕN at the top of the stanchion.The approach given by Davies (1990) will also be checked.

• The connections at the eaves, ridge and base will not be designed (see Chapter 7).• The background to the stanchion and rafter stability checks is given in Horne and

Ajmani (1971a, b), Horne et al. (1979).

EXAMPLE 8.3 Portal frame design.

Prepare a design in grade S355 steel for the frame whose basic geometry is given inFig. 8.15.

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Spacing 9000 ctrs.Purlin spacing 1250 ctrs (plan)

25000

5000

2200

FIGURE 8.15 Basic frame geometry for Example 8.3

Permanent loading (kPa): Sheeting 0,18Purlins 0,06Frame 0,18Total 0,42

UDL due to permanent loading: 0,42 × 9,0 = 3,78 kN/m

Nominal variable load (EN 1991-1-1): 0,60 kPa (or 5,40 kN/m)

Total factored design load: q = 1,35 × 3,78 + 1,5 × 5,40 = 13,20 kN/m

Since the purlins are at 1,25 m centres, it will be accurate enough to apply the roofloading as a UDL.

Preliminary design (geometry in Fig. 8.16):

Assume hh = 0,65 m;

h2 = hc − hh = 5,00 − 0,65 = 4,35 m

Distance from ridge to second purlin point, b′ = 2,5 m

l = L/2 − b′ = 12,5 − 2,5 = 10,0 m

Gradient of rafter, sr = 2,2/12,5 = 0,176

Height to second purlin point, r:

r = hc + lsr = 5,0 + 10 × 0,176 = 6,76 m

Moment at base of haunch, MB1:

MB1 = Hh2 = 4,35H (8.32)

Moment at second purlin point, M2:

The vertical reaction at the base of the stanchion V is given by

V = qL2

= 13,2252

= 165 kN

M2 = Vl − Hr − qL2

2= 10 × 165 − 6,76H − 13,2 × 102

2= 990 − 6,76H (8.33)

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h r

h 2h h

h

B

A

V

H

I

r

b�

L/2

2ndpurlin

Y

1stpurlin

X

L

b

c

Note : All distances to member cL’s unless otherwise indicated

FIGURE 8.16 Detail frame geometry

By setting M2 equal to Mb2 (i.e. equal strength rafter and column) Eqs (8.32) and(8.33) may be solved to give

H = Hbal = 89,1 kN

This will give equal plastic moment capacities in the column and rafter. This is not anoptimal solution as it is general practice to use a larger section column than rafter. Anoptimal solution can be obtained by increasing Hbal by around 15–20%. Increase Hbal

to 104 kN (i.e. an increase of 16,7%).

Thus from Eq. (8.32), MB1 (=Mpl,stanchion) is given as

MB1 = Hh2 = 4,35H = 4,35 × 104 = 452,4 kNm

Wpl = MplγM0

fy= 452,4 × 103 1,0

355= 1274 cm3

Select a 457 × 191 × 67 UKB (section classification Class 1: Mpl = 522 kNm)

and from Eq. (8.33) M2 (=Mpl,rafter):

M2 = 990 − 6,76H = 990 − 6,76 × 104 = 287 kNm

Wpl = MplγM0

fy= 287 × 103 1,0

355= 808 cm3

Select a 356 × 171 × 57 UKB (section classification is Class 1: Mpl = 359 kNm)

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Note both the rafter and stanchion have been overdesigned to allow for any reductionin carrying capacity due to sway classification and any slight changes in frame geometryover those assumed in the preliminary assessment of dimensions.

Check haunch depth:

h = 358 mm, so depth to base of haunch, hh = 1,5 × 358 = 537 mm

Hinge at B1 occurs at 5,00 − 0,537 = 4,463 m above base.

Thus H = Mpl,stanchion/h2 = 475/4,463 = 106,4 kN

The slight change in H can be ignored as it is small (2,3%).

Check load required to give collapse (q1) at the second purlin point, as the first hingeto occur is at the eaves (from an elastic distribution of bending moments).

Mpl,rafter = 10 × 12,5q1 − 106,4 × 6,76 − q1102

2= 75q1 − 719 (8.34)

Equate the value of Mpl,rafter for Eq. (8.34) with the actual value to give

75q1 − 719 = 326

or q1 = 13,9 kN/m

This is greater than the design load of 13,2 kN/m, thus collapse does not occur at thesecond purlin point.

Check moment at first purlin point from the apex:

l = 11,25 m; r = 5,00 + 0,176 × 11,25 = 6,98 m

M1 = Vl − Hr − ql2

2= 165 × 11,25 − 106,4 × 6,98 − 13,2 × 11,252

2= 278 kNm

This is less than Mpl,rafter, and is therefore satisfactory.

Check moment at end of haunch:

b = 2,5 + 0,5hstanchion = 2,5 + 0,5 × 0,4534 = 2,727 m

MH = Vb − (5,00 + 0,176b)H − qb2

2

= 165 × 2,727 − 106,4(5,00 + 0,176 × 2,727) − 13,2 × 2,7272

2= −182 kNm

Morris and Randall (1979) suggested this moment should be limited numerically to0,87Mpl,rafter/γ , where γ is the load factor between the total ultimate load and thecharacteristic loads. The 0,87 factor accounts for the approximate ratio between themoment to first yield and the plastic moment capacity (i.e. the shape factor).

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Total load factor, γ :

Load at ULS = 13,2 kN/m and total characteristic load = 3,78 + 5,40 = 9,18 kN/m, soγ = 13,2/9,18 = 1,44.

0,87Mpl,rafter

γ= 0,87 × 359

1,44= 217 kNm

An alternative is to calculate the moment capacity to first yield Myield divided by thetotal load factor:

Myield

γ= 1

γWel

fyγM0

= 11,44

8963551,0

× 10−3 = 221 kNm

Both methods give similar results, are less than the applied moment, and are thereforesatisfactory.

The maximum load the frame can carry qmax assuming hinges occur simultaneously atthe ridge and the base of the haunch is given by (Fig. 8.16):

qmax = 8L2

⎡⎣Mpl,R +

⎛⎝ 1 + hr

h

1 − hhh

⎞⎠Mpl,c

⎤⎦ (8.35)

or

qmax = 8L2

⎡⎣Mpl,R +

⎛⎝ 1 + hr

h

1 − hhh

⎞⎠Mpl,c

⎤⎦ = 8

252

⎡⎣359 +

⎛⎝ 1 + 2,2

5

1 − 0,5375

⎞⎠ 522

⎤⎦

= 15,37 kN/m

Thus the actual load factor γmax on the applied loading is

γmax = 15,379,18

= 1,67

The required load factor γ from above is 1,44.

Horizontal sway stability

The deflection δh under a force ϕN applied at the top of the (unhaunched) frame isgiven by

δh = φNh3

3EIc

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎛⎝ Irh

IcLr+ cos θ

IrhIcLr

⎞⎠⎛⎜⎜⎝1 −

2 IrhIcLr

+ 6 + 3 h1h

2(

IrhIcLr

+ 3 + 3 h1h +

(h1h

)2)

+ 2

⎛⎜⎜⎝ 2 Irh

IcLr+ 6 + 3 h1

h

4(

IrhIcLr

+ 3 + 3 h1h

(h1h

)2)⎞⎟⎟⎠

2⎞⎟⎟⎠−

0,25 −(

h1h

)2

IrhIcLr

cos θ

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(8.36)

The notation is given in Fig. 8.17

Ch08-H5060.tex 11/7/2007 14: 11 page 319


Lr

IrIr

Ic hIc

P � wN P � wN

FIGURE 8.17 Idealized frame for stanchion deflection

Calculation of ϕ from Eq. (8.16):

Determine αm from Eq. (8.18):

αm =√

0,5(

1 + 1m

)=√

0,5(

1 + 12

)= 0,866

Determine αh from Eq. (8.17) taking h as the height to eaves (as this is moreconservative)

αh = 2√h

= 2√5

= 0,894

The value αh is within the limits of 2/3 and 1,0, thus

φ = αmαhφ0 = 0,866 × 0,8941

200= 1

258

If N is taken as the axial load in the column, then it may be taken equal to V (=165 kN).

The length of the rafter Lr is given by

Lr = L2 cos θ

= 252 × cos 10

= 12,69 m

Equation (8.36) is best evaluated in sections:

IrhIcLr

= 16040 × 529380 × 12,69

= 0,215

2Irh

IcLr+ 6 + 3

h1

h= 2 × 0,215 + 6 + 3

2,25

= 7,75

IrhIcLr

+ 3 + 3h1

h+(

h1

h

)2

= 0,215 + 3 + 32,25

+(

2,25

)2

= 4,729

Ch08-H5060.tex 11/7/2007 14: 11 page 320


δh =1

258 165 × 53

3 × 210 × 106 × 29380 × 10−8

×((

0,215 + cos 100,215

)(1 − 7,75

2 × 4,729+ 2

(7,75

4 × 4,729

)2)

−0,25 −

(2,25

)2

0,215cos 10

⎞⎟⎠

= 1,133 × 10−3 m

Horne and Morris (1981) suggested that under a horizontal load of 1% of the totalapplied load the deflection should be limited to 0,0009hc. This is equivalent to0,00045hc under 0.5% (1/200) of the load applied to a single column. If the initialimperfection ϕ0 is modified by αmαh then the limiting deflection δlim is reduced to0,0045αmαhhc.

δlim = 0,00045αmαhhc = 0,00045 × 0,866 × 0,894 × 5 = 1,74 × 10−3 m

The actual deflection is lower and has been determined on an unhaunched frame.

An elastic analysis on the haunched frame gives a horizontal deflection of 0,823 mm,or h/6075 (=164,6 × 10−3 h). The deflection determined using the unhaunched frameis approximately 40% higher. This is acceptable.

Horne (1975) originally suggested that for multi-storey frames the elastic critical loadfactor αcr,H was given by

αcr,H = Hnom

δnom

hc

VULS(8.37)

where δnom is the horizontal deflection due to a nominal horizontal force Hnom, andVULS is the axial force in the column.

Thus using the results determined above for Hnom (as ϕN) and δnom as δh, αcr is given as

αcr,H =1

258 165

1,133 × 10−35

165= 17,1

The actual value of the notional horizontal load is not critical as only the load-deflectionratio is required, but clearly the applied load should not cause any type of failure.

However, Lim et al. (2005) have suggested that Horne’s approach to calculating αcr isunconservative for portal frames, that is, it overestimates αcr.

Lim et al. (2005) suggest a better estimate of αcr,est is given by

αcr,est = 0.8[

1 −(

NR,ULS

NR,cr

)max

]αcr,H (8.38)

Ch08-H5060.tex 11/7/2007 14: 11 page 321


where NR,ULS is the axial load in the rafter and NR,cr is the Euler buckling load ofthe rafter based on the span of the frame, and that the ratio should be taken at itsmaximum.

The maximum axial load in the rafter is at the haunch, and equals 133,5 kN.

Owing to the purlin restraints at 1,25 m centres means the rafter as a whole can onlybuckle about the major axis. The nominal buckling load Ncr assuming pinned endsand neglecting the effect of the haunches is given by

Ncr = π2EIr

L2 = π2210 × 106 × 29380 × 10−8

252 = 974,3 kN

Determine αcr,est from Eq. (8.38)

αcr,est = 0,8[

1 −(

NR,ULS

NR,cr

)max

]αcr,H = 0,8

[1 − 133,5

974,3

]17,1 = 11,8

From Lim et al. (2005), the second order plastic collapse load factor for a Category Atype frames, the second order plastic collapse load factor αp2 may be calculated usingthe Merchant–Rankine equation and the first order plastic collapse load factor αp1as

αp2 = αplαcr − 1

αcr(8.39)

From earlier the load factor αp1 (determined as γmax) is 1,67, thus

αp2 = αp1αcr − 1

αcr= 1,67

11,8 − 111,8

= 1,53

The required load factor on the frame is 1.44, thus the frame is satisfactory.

Check the method proposed by Davies (1990) for determining λcr:

λcr = 3EIr

Lr(hcNc,char + 0.3LrNr,char)(8.40)

where Nr,char and Nc,char are the forces in the rafter and column under characteristicframe loading, Lr is the rafter length, hc is the height to eaves, and Ic and Ir are themajor axis second moments of area of the column and rafter, respectively.

L = 25,0 mm; Lr = 12,69 m; hc = 5,0 m; Ic = 29380 cm4; Ir = 16040 cm4 (as only relativevalues are required).

Characteristic load/per unit run = 3,78 + 5,40 = 9,18 kN/m,

Nc,char = qL2

= 25 × 9,182

= 114,8 kN

To estimate Nr,char the following equation may be used

Nr,char = qL2(3 + 5m) cos θ

16Nhc+ 0,25qL sin θ (8.41)

Ch08-H5060.tex 11/7/2007 14: 11 page 322


where θ is the roof slope and the parameters N and m are given by

N = 2(

1 + m + m2 + Irhc

LrIc

)(8.42)

and

m = 1 + h1

hc(8.43)

where h1 is the rise from eaves to ridge.

m = 1 + h1

hc= 1 + 2,2

5= 1,44

N = 2(

1 + m + m2 + Irhc

LrIc

)= 2

(1 + 1,44 + 1,442 + 5 × 16040

12,69 × 29380

)= 9,457

Nr,char = qL2(3 + 5m) cos θ

16Nhc+ 0,25qL sin θ

= 9,18 × 252(3 + 5 × 1.44) cos 1016 × 9,475 × 5

+ 0,25 × 9,18 × 25 sin 10 = 86,0 kN

λcr = 3EIr

Lr(hcNc,char + 0,3LrNr,char)= 3 × 210 × 106 × 16040 × 10−8

12,69(5 × 114,8 + 0,3 × 12,69 × 86,0)= 8,83

This is a lesser value than that derived from the imposition of a 0,5% sway load.

Adopt the criterion in Horne and Merchant (1965) for the plastic load factor derivedfrom a modified Rankine type interaction equation with the lesser value of λcr.

As 4,6 < λcr < 10,0:

λp = 0,9λcr

λcr − 1= 0,9

8,838,83 − 1

= 1,015

Design UDL at ULS = 13,2 kN/m: actual carrying capacity = 15,37 kN/m (from above);surplus factor = 15,37/13,2 = 1,16. This is greater than that required by the Horne andMerchant check, so frame is satisfactory.

An alternative approach to checking stability is given in Horne and Morris (1981),where a span/depth ratio check is carried out. The notation has been slightly changedto give

L − bh

≤ 50W0

λWLhc

IcLrIrh

2 + IcLrIrh

240fy

cos θ (8.44)

W0 is the load required to cause collapse of the rafter assuming it is a straight mem-ber of exactly the same section as the original rafter. Any stiffening caused by theridge haunch can be ignored as it will have little effect. The eaves haunches cannotbe ignored, however. There are two possible collapse mechanisms to give W0 wherebya hinge always forms at the centre of the rafter beam, and the remaining two hinges

Ch08-H5060.tex 11/7/2007 14: 11 page 323


can occur either at the ends of the haunch in the parent rafter section, or at the endof the haunch in the enlarged section.

For the general case W0 is given by

W0 = 8(L − 2b)2 (Mpl,end + Mpl,centre) (8.45)

Where L is the span, b is the lengthy of the haunch (taken as zero for hinges formingadjacent to the stanchions), Mpl,end is the numerical value of the plastic moment cap-acity at the end of rafter and Mpl,centre that of the numerical value of that at the eaves.

Consider first the hinges at the haunch–rafter intersection:

In this case, b = 2,5 m and Mpl,end = Mpl,centre = 359 kNm, thus W0 is given by

W0 = 8(L − 2b)2 (Mpl,end + Mpl,centre) = 8

(25 − 2 × 2,5)2 (359 + 359) = 14, 36 kN/m

Now, consider hinges at the end of the haunch and the centre:

In this case, the hinge at the end of the haunch must occur in the weaker member,that is, the column with b = 0;

W0 = 8(L − 2b)2 (Mpl,end + Mpl,centre) = 8

252 (359 + 522) = 11,28 kN/m

Take the lesser value of W0, that is, 11,28 kN/m.

50W0

λWLhc

IcLrIrh

2+ IrLrIch

240fy

cos θ

= 50 × 11,2813,2

(255

)⎛⎝ 29380×12,695×16040

2 + 12,69×2938016040×5

⎞⎠(240

355

)cos 10 = 99,5

L − bh

= 25 − 2,50,358

= 62,9

The criterion is therefore satisfied, and the frame is stable.

Deflection check:

For an unhaunched frame with pinned feet under a UDL the vertical deflection δv atthe ridge is given by

δv = qL4

768EIr

⎡⎢⎢⎢⎣10 −

(8 + 5 h1

hc

) (3 + 2

(h1hc

))(

3+(

h1hc

)2 + 3 h1hc

+ IrhcIcLr

)2

×(

IrhIcLr

cos θ − 1cos θ

+ 3 +(

h1

hc

)2

+ 3h1

hc+ Irhc

IcLr

)⎤⎥⎥⎥⎦ (8.46)

Ch08-H5060.tex 11/7/2007 14: 11 page 324


In Eq. (8.46) the term IrhIcLr

cos θ−1cos θ

is negligible compared with the remainder of theterms in the last set of round parentheses, and thus Eq. (8.46) reduces to

δv = qL4

768EIr

⎡⎢⎢⎣10 −

(8 + 5 h1

hc

) (3 + 2

(h1hc

))(

3 +(

h1hc

)2 + 3 h1hc

+ IrhcIcLr

)⎤⎥⎥⎦

The horizontal deflection at the eaves δh is given by

δh = δv tan θ (8.47)

Determine δv and δh in terms of q:

δv = qL4

768EIr

⎡⎢⎣10 −

(8 + 5 h1

hc

) (3 + 2

(h1hc

))3+(

h1hc

)2 + 3 h1hc

+ IrhcIcLr

⎤⎥⎦

= 254q768 × 210 × 106 × 16040 × 10−8

⎡⎢⎣10 −

(8 + 5 2,2

5

) (3 + 2 2,2

5

)3 +

(2,25

)2 + 3 2,25 + 16040×5

29380×12,69

⎤⎥⎦

= 0,0246q

δh = δv tan θ = 0,0246q tan 10 = 4,34 × 10−3q

Vertical deflections:

Variable load of 5,4 kN/m:

δv = 0,0246q = 0,0246 × 5,4 = 0,133 m

This is equivalent to span/188.

Permanent load of 3,78 kN/m

δv = 0,0246q = 0,0246 × 3,78 = 0.093 m


Deflection under variable and permanent loading:

δv = 0,093 + 0,133 = 0,226 m


Horizontal deflections:

Variable load of 5,4 kN/m:

δh = 4,34 × 10−3q = 4,34 × 10−3 × 5,4 = 0.023 m

This is equivalent to height/217.

Ch08-H5060.tex 11/7/2007 14: 11 page 325


Permanent load of 3,78 kN/m

δh = 4,34 × 10−3q = 4,34 × 10−3 × 3,78 = 0,016 m

This is equivalent to height/313

Deflection under variable and permanent loading:

δh = 0,016 + 0,023 = 0,039 m

This is equivalent to height/128.

Woolcock and Kitipornchai (1986) recommended limits under service permanent load-ing of span/360, variable loading span/240 for vertical deflection and height/150 forhorizontal deflection. It appears unclear which loads are to be used for horizontaldeflections.

The horizontal deflection under variable and permanent loading marginally exceedstheir recommendation but the vertical deflection well exceeds them. However, itshould be noted that the frame is stiffer than assumed in the calculations owing tothe presence of haunches and the stiffness of the roof sheeting has been ignored. Thusthe vertical deflection calculated above will be accepted as satisfactory.

A computer analysis allowing for the haunches gives a vertical deflection at the eavesunder

Variable load: 79,2 mm (or span/316)

Permanent plus variable load: 134,7 mm (or span/186)

The analysis without haunches overpredicts the vertical deflections by almost 70%.Thus the estimates with no allowance for haunches are extremely conservative. Thedeflections determined with haunches will be reduced even further if the effects ofcladding are considered.

Column stability:

There are two possible approaches, the first is to supply a torsional restraint at adistance Lm below the hinge position and then to check the remainder of the columnbelow this restraint for the combined effects of strut buckling and lateral torsionalbuckling. The second is to check the whole length of the column below the hingemaking use of restraint to the tension flange by the sheeting rails.

Method 1

The maximum distance to point of restraint in a column Lm is given by

Lm = 38iz√1

57.4NEd

A + 1756C2

1

W 2pl,y

AIt

( fy235

)2(8.48)

Ch08-H5060.tex 11/7/2007 14: 11 page 326


Note, Eq. (8.48) must be solved iteratively as the moment gradient factor C1 isdependent upon the value of the bending moment at a distance Lm below the hinge.A conservative solution may be obtained by setting C1 = 1.0 (uniform moment).

As an alternative, the length to the restraint may be taken equal to Ls where Ls isgiven by

Ls = Lk√

CmMpl,y,Rk

MN,y,Rk + aNEd(8.49)

where Cm is the modification factor for a linear moment gradient, MN,y,Rd is themoment capacity of the column reduced due to axial load and Lk is given

Lk =htf

(5,4 + 600fy

E

)iz√

5,4fyE

(htf

)2 − 1

(8.50)

Use Eq. (8.49):

Force in the column (NEd) = 165 kN;

NEd

A= 165 × 103

8550= 19,3 MPa

Suitable sheeting rails for this frame are 202 mm deep, thus

a = 0,5(453,4 + 202) = 327,7 mm

Determine MN,y,Rk:

Determine the ratio n between the applied force NEd and the plastic axial resistanceNpl,Rd

n = NEd

Npl,Rd= 165 × 103

8550 3551,0

= 0,054

Determine the ratio a between the total web area (excluding just the flanges) and thecross-section area:

a = A − 2btfA

= 8550 − 2 × 189,9 × 12,78550

= 0,436

MN,y,Rd = Mpl,y,Rd1 − n

1 − 0,5a= Mpl,y,Rd

1 − 0,0541 − 0,5 × 0,436

= 1,21 Mpl,y,Rd

Thus MN,y,Rd = Mpl,y,Rd = 522 kNm.

Lm = 38iz√1

57,4NEd

A + 1756C2

1

W 2pl,y

AIt

( fy235

)2

= 38 × 41,2√1

57,4 19,3 + 1C2

1

(1471×103)2

756×8550×37,1×104

(355235

)2= 1566√

0,336 + 2,059C2

1

Ch08-H5060.tex 11/7/2007 14: 11 page 327


1137

4467

A

(354 kNm)

522,4 kNmUndersideof haunch

Position ofrestraint

FIGURE 8.18 BMD instanchion AB

Use Eq. (5.40) to determine C1. Under a constant moment gradient, C1 = 1,0, thusLm = 1012 mm. After three iterations, ψ = 0,678, C1 = 1,148 and Lm = 1137 mm.

Check column buckling below this level under axial load and moment (Fig. 8.18):

Mz,Ed = 0: NSd = 165 kN:

My,Ed = 106,4 × (4,463 − 1,137) = 354 kNm.


Flanges: Class 1.

Web:

Actual slenderness:ctw

= dtw

= 407,68,5

= 48,0

Length of web xw required to carry axial force:

xweb = NEd

twfy

γM0

= 165 × 103

8,5 3551,0

= 54,7 mm

Depth of web in compression αc:

αc = d2

+ xweb

2= 407,6 + 54,7

2= 231,2 mm

or

α = (αc)c

= 231,2407,6

= 0,567

Limit for Class 1, with α > 0,5:

dtw

=396√

235fy

13α − 1=

396√

235355

13 × 0,567 − 1= 50,6

Ch08-H5060.tex 11/7/2007 14: 11 page 328


The actual web slenderness is below this, therefore section is Class 1.

As the compression flange is unrestrained, the lateral torsional buckling must bechecked.

As the section is Class 1, Eqs (6.61) and (6.62) of EN 1993-1-1 may be simplified as�My,Ed and �My,Ed are both zero, as is Mz,Ed to give

NEd

χyAfy

γM1

+ kyyMy,Ed

χLTWpl,yfy

γM1

≤ 1,0

NEd

χzAfy

γM1

+ kzyMy,Ed

χLTWpl,yfy

γM1

≤ 1,0

Buckling curves:

hb

= 453,4189,9

= 2.39 > 1.2

y–y axis curve ‘a’ (α = 0.21) and z–z axis curve ‘b’ (α = 0,34).

Calculation of χy:

For yy axis buckling, the system length is taken as base of the haunch and the bottomof the column as restraints are present at both these points, that is, a length of 4,463 m

Ncr,y = π2EIy

L2y

= π2 × 210 × 106 × 29380 × 10−8

4,4632 = 30572 kN

λ̄y =√

Af y

Ncr,y=√

8550 × 35530572 × 103 = 0,315

�y = 0,5[1 + α(λ̄y − 0,2) + (λ̄y)2] = 0,5[1 + 0.21(0,315 − 0,2) + 0,3152] = 0,562

χy = 1

�y +√

�2y − (λ̄y

)2 = 1

0,562 +√0,5622 − 0,3152= 0,973

Calculation of χz:

For yy axis buckling, the system length is taken between the restraint 1,137 m below thehaunch and the bottom of the column as restraints are present at both these points,that is, a length of 3,362 m

Ncr,z = π2EIz

L2z

= π2 × 210 × 106 × 1452 × 10−8

3,3622 = 2663 kN

λ̄z =√

Af y

Ncr,y=√

8550 × 3552663 × 103 = 1,064

�z = 0,5[1 + α(λ̄z − 0,2) + (λ̄z)2] = 0,5[1 + 0,34(1,064 − 0,2) + 1,0642] = 1,213

χz = 1

�z +√

�2z − (λ̄z

)2 = 1

1,213 +√1,2132 − 1,0642= 0,567

Ch08-H5060.tex 11/7/2007 14: 11 page 329


Calculation of χLT:

The system length for lateral torsional buckling is taken as the distance between therestraint 1,137 m below the haunch and the bottom of the column, that is, a length of3,362 m

Determine Mcr from Eq. (5.5)

Iw

Iz= 0,705 × 10−6

1452 × 10−8 = 0,0486 m2

π2EIz

L2 = π2 × 210 × 106 × 1452 × 10−8

3,3622 = 2663 kN

L2GI t

π2EIz= 81 × 106 × 37,1 × 10−8

2663= 0,0113 m2

Mcr = π2EIz

L2

[Iw

Iz+ L2GI t

π2EIz

]1/2

= 2663[0,0486 + 0,0113]1/2 = 652 kNm

Determine C1 from Eq. (5.40):

ψ = 0,1

C1= 0,6 + 0,4ψ = 0,6 + 0,4 × 0 = 0,6

or, C1 = 1,667.

From Eq. (5.12) with Mcr modified by C1,

λ̄LT =√

Wpl fyC1Mcr

=√

1471 × 103 × 355 × 10−6

1,667 × 652= 0,693

Use the general case for lateral torsion buckling, as h/b > 2, αLT = 0,34 (from Table 5.1)

�LT = 0,5[1 + αLT(λ̄LT − 0,2) + (λ̄LT)2] = 0,5[1 + 0,34(0,693 − 0,2) + 0,6932]

= 0,824

χLT = 1

�LT +√

�2LT − (λ̄LT

)2 = 1

0,824 +√0,8242 − 0,6932= 0,788

Use Annex B of EN 1993-1-1 to determine kyy and kzy:

As the moment gradient is linear with the least value of moment equal to zero, ψ = 0,so from Table B.3, all the Cm values are 0,6.

kyy = Cmy

⎛⎝1 + (λ̄y − 0,2)

NEd

χyAfy

γM1

⎞⎠ ≤ Cmy

⎛⎝1 + 0,8

NEd

χyAfy

γM1

⎞⎠

= 0,6

⎛⎝1 + (0,315 − 0,2)

165 × 103

0,973 × 8550 3551,0

⎞⎠ ≤ 0,6

⎛⎝1 + 0,8

165 × 103

0,973 × 8550 3551,0

⎞⎠

= 0,604 ≤ 0,627

Ch08-H5060.tex 11/7/2007 14: 11 page 330


Thus, kyy = 0,604.

As torsional deformations can occur (although they will be in practice slightly limitedby the sheeting rails)

kzy = 1 − 0,1λ̄z

CmLT − 0,25NEd

χzAfy

γM1

≥ 1 − 0,1CmLT − 0,25

NEd

χzAfy

γM1

= 1 − 0,1 × 1,0640,6 − 0,25

165 × 103

0,567 × 8550 3551,0

≥ 1 − 0,10,6 − 0,25

165 × 103

0,567 × 8550 3551,0

= 0,971 ≥ 0,973

kxy = 0,973

NEd

χyAfy

γM1

+ kyyMy,Ed

χLTWpl,yfy

γM1

= 165 × 103

0,973 × 8550 3551,0

+ 0,604354 × 106

0,788 × 1471 × 103 3551,0

= 0,576 ≤ 1,0

NEd

χzAfy

γM1

+ kzyMy,Ed

χLTWpl,yfy

γM1

= 165 × 103

0,567 × 8550 3551,0

+ 0,973354 × 106

0,788 × 1471 × 103 3551,0

= 0,942 ≤ 1,0

Thus the column is satisfactory below the torsional restraint.

Method 2

Determine the elastic critical torsional buckling load. This is given by Eq. (5.100)together with an additional term π2EIza2/L2

t to allow for the restraint from thesheeting rails.

a = 327,7 mm (from Method 1)

NcrT = 1i2s

(π2EIza2

L2t

+ π2EIw

L2t

+ GIt

)

where Lt is the distance between restraints to both flanges. Try Lt as the height fromthe underside of the haunch to the base, that is, 4,463 m, and is from (Eq. (5.101))modified by an additional term a2 to give

i2s = i2z + i2y + a2

or

i2s = i2z + i2y + a2 = (41,22 + 1852 + 327,72) × 10−6 = 0,1433 m2

Ch08-H5060.tex 11/7/2007 14: 11 page 331


or, is = 379 mm.

NcrT = 1i2s

(π2EIza2

L2t

+ π2EIw

L2t

+ GIt

)

= 10,1433

(π2 × 210 × 106 × 1452 × 10−8

4,4632

+π2 × 210 × 106 × 0,705 × 10−6

4,4632 + 81 × 106 × 37,1 × 10−8

)

= 11265 kN

NcrE = π2EIz

L2t

= π2 × 210 × 106 × 1452 × 10−8

4,4632 = 1511 kN

η = NcrE

NcrT= 1511

11265= 0,134

B0 = 1 + 10η

1 + 20η= 1 + 10 × 0,134

1 + 20 × 0,134= 0,636

B1 = 5√

η

π + 10√

η= 5

√0,134

π + 10√

0,134= 0,269

B2 = 0,51 + π

√η

− 0,51 + 20η

= 0,51 + π

√0,134

− 0,51 + 20 × 0,134

= 0,097

Determine the ratio βt defined as the ratio of the algebraically smaller end momentto the larger end moment. As the smaller end moment (at the base) is zero, βt = 0.

Cm = 1B0 + B1βt + B2β

2t

(8.51)

As βt = 0, Eq. (8.51) reduces to

Cm = 1B0

= 10,636

= 1,572

Lk is given by Eq. (8.50):

Lk = iz

htf

(5,4 + 600fy

E

)√

5,4fyE

(htf

)2 − 1

= 41,2

453,412,7

(5,4 + 600×355

210×103

)√

5,4 355210×103

(453,412,7

)2 − 1

= 2893 mm

The reduced moment capacity MN,y,Rk due to the axial load equals the plastic momentcapacity Mpl,Rk.

Ls =√

CmLk

(Mpl,y,Rk

Mn,y,Rk + aNEd

)= 2893

√1,572

(522

522 + 165 × 0,3277

)= 3286 mm

This is less than the height to the underside of the haunch, and slightly less than theheight to the restraint of 3362 mm.

Ch08-H5060.tex 11/7/2007 14: 11 page 332


Recheck using Lt = 3362 mm.

NcrT = 1i2s

(π2EIza2

L2t

+ π2EIw

L2t

+ GIt

)

= 10,1433

(π2 × 210 × 106 × 1452 × 10−8

3,3622

+ π2 × 210 × 106 × 0,705 × 10−6

3,3622 + 81 × 106 × 37,1 × 10−8

)

= 19692 kN

NcrE = π2EIz

L2t

= π2 × 210 × 106 × 1452 × 10−8

3,3622 = 2663 kN

η = NcrE

NcrT= 2663

19692= 0,135

B0 = 1 + 10η

1 + 20η= 1 + 10 × 0,135

1 + 20 × 0,135= 0,635

Hence, Lk = 3290 mm. This is probably adequate.

Rafter

(a) Haunch between connection and bracing at first purlin point.

Relevant dimensions are given in Fig. 8.19.

227

227 2500

538

179

‘zed’ purlins at 1250 ctrs

FIGURE 8.19 Haunch geometry

Ch08-H5060.tex 11/7/2007 14: 11 page 333


At any point along the rafter the value of the applied moment MSd is given by

MEd,x = 165x − 106,4(5 + 0,176x) − 13,2x2

2At the connection, x = 0,227 m (ignoring end plate) and

MSd = −499 kNm.

At the restraint x = 1,25 m, MSd = −359,5 kNm.

Section classification (with axial force)

Flanges are Class 1.

Web:

Determine the axial force in the member:

NSd = H cos 10◦ + V sin 10◦ = 106,4 cos 10 + 165 sin 10 = 133,4 kN.

Web classification check is not needed as the web is restrained from buckling due tothe end plate welded to the web and flanges forming the end plate for the connection.

Determine section properties ignoring both the central flange of the haunch and thefillets.

A = 2btf + dtw = 2 × 0,1722 × 0,013 + 0,0081(0,716 − 2 × 0,013) = 0,010066 m2

Iy = 112

[bh3 − (b − tw) (h − 2tf )3

]

= 112

[0,1722 × 0,7163 − (0,1722 − 0,0081) (0,716 − 2 × 0,013)3]

= 0,775 × 10−3 m4

Iz = 112

[2tfb3 + (h − 2tf )t3

w

]

= 112

[2 × 0,013 × 0,17223 + (0,716 − 2 × 0,013)0,00813

]= 11,09 × 10−6 m4

Plastic section modulus, Wpl,y:

Wpl,y = 14

[bh2 − (b − tw)(h − 2tf )2

]

= 14

[0,1722 × 0,7162 − (0,1722 − 0,0081) (0,716 − 2 × 0,013)2

]= 2,538 × 10−3 m3

Warping constant, Iw:

Iw = Izh2s

4= 11,09 × 10−6(0,716 − 0,013)2

4= 1,370 × 10−6 m6

where hs is the depth between the centroids of the flanges.

Ch08-H5060.tex 11/7/2007 14: 11 page 334


To determine the torsional constant, It, it will be sufficiently accurate to treat theflanges and webs as thin, thus,

It = 13

∑bt3 = 1

3

[2 × 0,1722 × 0,0133 + (0,716 − 2 × 0,013) 0,00813

]= 0,374 × 10−6 m4

The distance between restraints, L, is the slope distance between the end of the rafterand the first purlin, and is given by

L = 1,250 − 0,277cos 10◦ = 0,988 m

Determine Mcr from Eq. (5.5)

Iw

Iz= 1,370 × 10−6

11,09 × 10−6 = 0,1235 m2

π2EIz

L2 = π2 × 210 × 106 × 11,09 × 10−6

0,9882 = 23547 kN

L2GIt

π2EIz= 81 × 106 × 0,374 × 10−6

23547= 1,287 × 10−3 m2

Mcr = π2EIz

L2

[Iw

Iz+ L2GIt

π2EIz

]1/2

= 23547[0,1235 + 1,287 × 10−3]1/2

= 8318 kNm

Determine C1 from Eq. (5.40),

ψ = 359,5492

= 0,731

1C1

= 0,6 + 0,4ψ = 0,6 + 0,4 × 0,731 = 0,892

or, C1 = 1,121.

λLT =√

Wpl fyC1Mcr

=√

2,538 × 10−3 × 3551,121 × 8318 × 10−3 = 0,311

As λLT < 0,4, lateral torsional buckling cannot occur.

Haunch instability

An alternative approach is to calculate the limiting length between lateral torsionrestraints (i.e. between the haunch and the purlin immediately after the end of thehaunch) (cl BB.3).

Ls = Lk√

Cn

c(8.52)

Ch08-H5060.tex 11/7/2007 14: 11 page 335


where Lk is the basic critical length, and c is a factor allowing for the shape of thehaunch given by

c = 1 + 3htf

− 9

(hh

hs

) 23√

Lh

Ly(8.53)

where h/tf is the ratio of the depth to flange thickness tf for the basic section, LhLy isthe proportion of the length between the restraints taken up by the taper, and hh/hs

is the ratio of additional depth of the beam at the haunch to that of the basic section.The basic length Lk is given by Eq. (8.50).

Determine Lh/Ly:

Length of taper 2500 mm, and horizontal distance from inside of stanchion to nextpurlin beyond end of taper is 3 × 1250 − 227 = 3523 mm, so

Lh

Ly= 2500

3523= 0,710

As the haunch is fabricated using the same section size as the parent section minusone flange, hh = 358 − 13 = 345 mm, so

hh

hs= 345

358= 0,964

c = 1 + 3htf

− 9

(hh

hs

)23√

Lh

Ly= 1 + 3

35813 − 9

0,96423√

0,710 = 1,133

Determine Lk from Eq. (8.50):

Lk = izhtf

5,4 + 600fyE[

5,4fyE

(htf

)2 − 1]1/2 = 39,1

35813

5,4 + 600 355210×103[

5,4 355210×103

(35813

)2 − 1]1/2 = 2838 mm

Determination of Cn:

To determine Cn values of the parameter R are required which is defined by

R = My,Ed + aNEd

Wpl,y, fy

From safe load tables a suitable depth purlin is 232 mm, thus the depth a between thecentroid of the purlins and the centroid of the member is given by

a = 0,5h + 0,5 × 232

Take the value of NEd as at the end of the member, that is, 134,4 kN.

The values of R are determined in Table 8.1

From Table 8.1 the maximum value of R, RS = 0,633: RE is max (R1, R5) = 0,627.

Ch08-H5060.tex 11/7/2007 14: 11 page 336


Cn = 12R1 + 3R2 + 4R3 + 3R4 + R5 + 2(RS − RE)

RS − RE = 0,633 − 0,627 = 0,006 > 0

So,

Cn = 12R1 + 3R2 + 4R3 + 3R4 + R5 + 2(RS − RE)

= 120,625 + 3 × 0,629 + 4 × 0,632 + 3 × 0,633 + 0,627 + 2 × 0,006

= 1,584

Ls = Lk√

Cn

c= 2838

√1,584

1,13= 3158 mm

This exceeds the length of the haunch so there need only be restraints at haunchconnection and the end of the haunch.

Check between haunch and first purlin point beyond point of contraflexure.

Point of contraflexure occurs when M = 0, or

165x − 106,4(5 + 0,176x) − 6,6x2 = 0, or x = 4,584 m

The distance from the end of the haunch to the point of contraflexure is4,584 − 2,727 = 1,857 m.

The distance to the next purlin point is 5,00 − 2,727 = 2,273 m.

This is less than the value of Lk and therefore no additional restraint is required.

(c) Ridge

As the moment gradient is non-linear, the stable length is given by Lk = 2833 mm(calculated above). This is greater than the slope length between purlins.

8.14.2 Notes on the Design of a Gable End Frame

• The rafter is usually designed as a continuous beam spanning over the gable endcolumns, with simple, non-moment transferring connections to the frame. This will

TABLE 8.1 Calculation of the values of R for the haunch.

1 2 3 4 5

Distance x (mm) 227 852 1477 2102 2727Depth h (mm) 716 626,5 537 447,5 358,0A (mm2) 10156 9431 8706 7981 7256I y (×108 mm4) 7,854 5,556 4,031 2,658 1,603W pl (×106 mm3) 2,538 2,104 1,702 1,332 0,996Mx (kNm) 499,1 412,2 330,4 253,7 182,2aN Ed (kNm) 63,7 57,7 51,7 45,7 39,6R 0,625 0,629 0,632 0,633 0,627

Ch08-H5060.tex 11/7/2007 14: 11 page 337


mean that if the spans are equal, the first bay will be critical since it will only becontinuous at one end.

• The centre lines of the gable end columns should be coincident with the spacing ofthe roof purlins in order to resist the reaction from the wind loading on the gableend columns without subjecting the rafters to bi-axial bending. It is also preferableif the spacing of the gable end columns can be similar to the frame spacing in orderto keep the same section for the sheeting rails, although this is not essential.

• Often the loading on the gable end will require the use of only relatively smallsections, but detailing requirements and compatibility with the rest of the framemay make it necessary to use larger sections.

8.14.3 Multi-bay portal frames

It is not intended to present a design example of such a frame since the principles andmethods are similar to those for single bay frames. It is intended, rather, to point outadditional factors that need to be taken into account.

• The critical bay, assuming as in normal practice, that all the spans are equal, is theend bay. However, it must be noted that it is no longer possible to assume uniformsnow loading. The frame should be first checked under the nominal roof loadingof 0,6 kPa (modified for slope if necessary) as a variable load together with thepermanent loading. It should then be checked for the accidental snow drift loadingtogether with the permanent loading considered as an accidental load combinationwith reduced partial safety factors.

(b) Exterior bays after snap through

(a) Interior bays before snap through

FIGURE 8.20 Snap through failure of multi-bay portal frames

Ch08-H5060.tex 11/7/2007 14: 11 page 338


• Although with balanced loading the interior stanchions carry axial load only, andthe internal rafters have a higher reserve of strength, it is normal practice to detailall bays identically. This is partly to achieve fabrication economies, and to avoidproblems on site with varying rafter sizes, and partly to avoid deflection and ‘snapthrough’ failure.

• Sway stability is either handled by the imposition of a nominal sway force as forsingle bay frames, or if the empirical method of Horne and Morris (1981) is usedas in Eq. (8.44), then the ratio of the column stiffness (Ic/hc) to the rafter stiffness(Ir/L) should be halved.

• In frames with more than two bays it is possible for snap through to occur wheninstability happens as the top of the stanchions spread followed by inversion of therafter (Fig. 8.20)

This can be checked using the following empirical equation (Horne and Morris, 1981)

L − bh

=25(

4 + Lhc

) (1 + Ic

Ir

)λWW0

(λWW0

− 1) 240

fytan 2θ (8.54)

where symbols are defined in Eq. (8.44). If the arching ratio is less than one, snapthrough cannot occur, so there is no limit on the effective length to depth ratio.

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Ch09-H5060.tex 12/7/2007 14: 9 page 341

C h a p t e r 9 / Trusses

This chapter is concerned with the design of triangulated and non-triangulated trusses.With the advent of rolled hollow sections the design of both types of truss has beenrevolutionized.

The use of rolled hollow sections for trusses provides a far more efficient use ofmaterial as the buckling strengths are higher as radii of gyration are larger and lateraltorsional buckling is either non-existent in the case of square or circular sections, orthe effects are much reduced for rectangular sections. Rolled section trusses are mucheasier to fabricate as the welding is generally fillet welds or full strength butt welds.However, it needs remembering that room must be provided to allow the welds to beproperly formed. This can be achieved by limiting the number of members at a nodeto the main chord member and at most two subsidiary web members and limiting theminimum angle between members to 30◦. A problem however with rolled sections isforming joints to enable large trusses to be transported. If circular sections are usedas main chord members fabrication problems and resultant high costs may ensue.Maintenance, and painting, is much easier with rolled sections as spray techniquescan be employed. Note, rolled hollow sections are commonly available in Grade S355,but it may be difficult to obtain such sections in S275.

9.1 TRIANGULATED TRUSSES

It is generally acceptable to analyze triangulated trusses on the basis of pinned joints atthe nodes together with loading from the roof system applied at nodes. However, wherepurlins exist between the nodes then the relevant members need to designed underthe effects of flexure form the loads concerned. It is conservative to consider the mainchord members as simply supported between nodes, although use of continuity can bemade. Where it is necessary to consider the existence of services within the roof space,such loading can be considered as nodal loads on the bottom chord of the truss. Theeffect of service loads must be neglected when considering the effects of wind uplift.

It is not generally necessary to consider the effects of secondary moments due to jointfixity as these are usually low. Note, however if the original pin-jointed analysis isperformed using a plane frame computer analysis package then it is relatively easy to

Ch09-H5060.tex 12/7/2007 14: 9 page 342

342 • Chapter 9 / Trusses

check the effect of secondary moments by using the sections determined as suitable tocarry the pinned forces and simply altering the end conditions of the members. Also,provided the secondary moments are negligible, the deflections can be determinedfrom a pin-joint analysis. It should be noted that the actual deflections in a truss willbe less than the calculated deflections as any stiffening effect of the roof claddingand purlins has been ignored. Note that even if the applied actions due to wind upliftand permanent loads are lower than those due to wind downthrust, roof loading andpermanent loads, consideration still needs to be given to bracing on the lower boomof the truss as this is now in compression. This will almost certainly mean the provisionof longitudinal bracing to ensure system lengths are kept reasonably small to ensurebuckling does not occur out of plane.

EXAMPLE 9.1 Triangulated truss design.

Prepare a design using square and rectangular rolled hollow sections in Grade S355steel for the truss whose geometry is given in Fig. 9.1.

The load due to initial imperfections has been omitted partly because the value of theload is small and partly because the truss is triangulated and thus will exhibit virtuallyno sway effects.

Actions (kPa):

Permanent: Sheeting 0,22Purlins 0,08Total 0,30

UDL for each truss = 0,30 × 9,0 = 2,7 kN/m.

Assume self-weight is 1,5 kN/m, thus total permanent action is 4,2 kN/m.

Since the purlins are at nodal points, nodal permanent action = 3,0 × 4, 2 = 12,60 kN(on an internal node).

Variable action:

From EN 1991-1-1 variable action (with no access, snow drift ignored) is 0,6 kPa, sonodal load = 0,6 × 9,0 × 3,0 = 16,2 kN

A

BC

FH J

LN

PR

D E G I K M O Q

8 bays @ 3 m

3,2

m

0,8

m

FIGURE 9.1 Truss geometry for EXAMPLE 9.1

Ch09-H5060.tex 12/7/2007 14: 9 page 343


Although wind loading will need considering as it may cause uplift due to suction, in thisexample only permanent and variable actions due to roof loading will be considered.Thus the only combination of actions needed is 1,35gk + 1,5qk.

So the total nodal action is 1,35 × 12,6 + 1,5 × 16,2, that is, 41,31 kN per internal node.

The truss will be designed using a pin-joint analysis, and then checked using a fixed-joint analysis. The effect of notional horizontal loading will be ignored for this designexample as its effect will be negligible.

Table 9.1 shows the results from a pin-joint analysis under factored variable andpermanent actions for half the truss.

Member design:

The top and bottom chord together with the end posts will be fabricated from thesame section.

Maximum force (compressive) NSd is in member CF and is 379 kN.

(The maximum tensile force is 320 kN and will not be critical.)

From cl BB.1.3 the buckling length may be taken as 0,9L in both planes where L is thesystem length.

Effective length:

LFC = 3,105 m (slope length).

TABLE 9.1 Member forces for EXAMPLE 9.1.

Fixed-joint analysis

Pin-joint analysis MomentAxial force Axial force Shear force End 1 End 2

Member kN kN kN kNm kNm

AB −165,24 −163,38 −7,19 2,85 −2,93CD −82,62 −82,12 −1,57 −1,07 1,16EF −28,92 −29,33 0,04 0,04 −0,04GH 9,54 9,17 0,12 −0,12 0,17IJ 82,62 81,63 0 0 0AD 0 7,19 1,86 −2,85 2,77DE 309,83 307,14 −0,50 1,90 0,87EG 371,79 371,29 0,04 0,50 0,62GI 357,50 356,79 −0,37 0,83 0,29BC −315,98 −311,97 1,82 0,29 −0,29CF −379,14 −378,60 −0,37 −1,61 −0,45FH −364,56 −364,02 0,12 −0,45 −0,78HJ −315,98 −316,14 −0,54 0,66 −0,99BD 320,64 308,46 0,17 0,29 −0,29CE 68,37 73,28 −0,08 0,04 0,25FG −17,18 −17,27 0 0,04 0,08HI −63,08 −61,76 0 0,04 0

Ch09-H5060.tex 12/7/2007 14: 9 page 344


Out-of-plane buckling:

Buckling effective length is the length of the member, that is, 3,105 m.

In-plane buckling

Buckling length is 0,9LFC = 2,795 m, as there is a large degree of restraint owing tothe welded joints (and continuity of the chord member).

Try a 100 × 100 × 8 HFRHS.

A = 2910 mm2, i = 37,4 mm, I = 4,08 × 10−6 m4.


ct

= b − 2r − 2tt

= 100 − 2 × 8 − 2 × 88

= 8,5

Limit for Class 1:

33ε = 33

√235fy

= 33

√235355

= 26,8

Section is Class 1.

Use the greatest system length of 3,105 to calculate Ncr:

Ncr = π2EIL2 = π2 × 210 × 106 × 4,08 × 10−6

3,1052 = 877 kN

λ =√

AfyNcr

=√

2910 × 355 × 10−3

877= 1,085

For a rolled hollow section, buckling curve ‘a’ is used, α = 0,21:

� = 0,5 [1 + α(λ − 0,2) + (λ)2] = 0,5 [1 + 0,21(1,085 − 0,2) + 1,0852] = 1,182

χ = 1

� +√

�2 − (λ)2= 1

1,182 +√1,1822 − 1,0852= 0,606

Nb,Rd = χAfy

γM1= 0,606 × 2910

3551,0

× 10−3 = 626 kN

All the remaining members will be the same section size in order to aid fabrication.

Maximum force NSd is in member BC and is 320 kN (tension).

Try a 60 × 60 × 6,3 HFRHS.

A = 1330 mm2, i = 21,7 mm, I = 63,4 × 108 mm4.


ct

= b − 2r − 2tt

= 60 − 2 × 6,3 − 2 × 6,36,3

= 5,52

Ch09-H5060.tex 12/7/2007 14: 9 page 345


Limiting value for Class 1:

ct

= 33

√235fy

= 33

√235355

= 26,8.

Section is therefore Class 1.

Nt,Rd = Afy

γM0= 1330

3551,0

× 10−3 = 472 kN

However the compression members also need checking.

Member DC carries the largest compression force of −82,6 kN.

Length of DC is 1,4 m.

Critical effective length is that for out-of-plane buckling, that is, 1,4 m.

Ncr = π2EIL2 = π2 × 210 × 106 × 63,4 × 10−8

1,42 = 670 kN

λ =√

AfyNcr

=√

1330 × 355670 × 103 = 0,839

For a hot rolled RHS, α = 0,21, so

� = 0,5 [1 + α(λ − 0,2) + (λ)2] = 0,5 [1 + 0,21(0,839 − 0,2) + 0, 8392] = 0,987

χ = 1

� +√

�2 − (λ)2= 1

0,987 +√0,9872 − 0,8392= 0,664

Nb,Rd = χAfy

γM1= 0,664 × 1330

3551,0

× 10−3 = 314 kN > NSd

The deflections were also assessed on a pin-jointed analysis and gave 0,018 munder variable load (span/1333) and 0,033 m (span/727) under variable together withpermanent loading. Both these deflection ratios are acceptable.

Check on self-weight of the truss:

From the member sizes determined on the basis of the force distribution for thepin-joint analysis, the total mass of the truss was 1,6 ton compared with the estimateof 3,6 ton. This means that both the total forces and the total deflection have beenoverestimated by around 10%. The check on self-weight is carried out in Table 9.2.

Web capacity in the bottom chord at A (Section 4.8).

The reaction is 165,2 kN.

Note as a rolled hollow section is being used, fyw = fyf = 355 MPa. Since the length ofstiff bearing ss is not known, set ss = 0 and determine the value required should thecheck fail.


Ch09-H5060.tex 12/7/2007 14: 9 page 346


TABLE 9.2 Check on self-weight estimate for EXAMPLE 9.1.

Member Total length (m) Sub total (m) Mass/unit length (kg/m) Mass (kg)

AQ 24AB and QR 1,6BJ and JR 24,48 50,08 22,9 1147BC and OR 6,21DE and PH 6,62FG and NK 7,21HI and LI 7,68DC and PO 2,8EF and NH 4,0HG and LK 5,2JI 3,2 42,92 10,5 451Total mass 1598

Determine m1:

m1 = fyfbf

fywtw= bf

tw= 50

8= 6,25


As ss and c have been assumed to be zero, then lc = 0, then the least value of ly isgiven by

ly = tf

√m1

2= 8,0

√6,25

2= 14,1 mm

The depth of the web hw has been taken as the clear depth of the web,

hw = h − 2r − 2t = h − 4t = 100 − 4 × 8 = 68 mm

FCR = 0,9kFEt3w

hw= 0,9 × 2 × 210

8,03

68= 2846 kN

λF =√

lytwfyw

FCR=√

14,1 × 8,0 × 3552846 × 103 = 0, 119

As λF < 0,5, m2 = 0

χF = 0,5λF

= 0,50,119

= 4,2


Leff = χFly = 1,0 × 14,1 = 14,1 mm

FRd = Leff twfyw

γM1= 14,1 × 8,0

3551,0

× 10−3 = 40 kN

The total resistance therefore is 80 kN (from both webs).

A stiff bearing is therefore required to supply a further 165,2 − 80 = 85,2 kN.

Ch09-H5060.tex 12/7/2007 14: 9 page 347


Try a 25 mm stiff bearing.

kF = 2 + 6(

ss + chw

)= 2 + 6

(25 + 0

68

)= 4,21

lc = kFEt2w

2fywhw≤ ss + c = 3,33 × 210 × 103 × 82

2 × 355 × 68≤ 25 + 0 = 927 ≤ 25

The limiting value of lc is greater than ss, so lc = 25 mm.

ly = lc + tf

√m1

2+(

lctf

)2

= 25 + 8,0

√6, 25

2+(

258

)2

= 53,7 mm (a)

ly = lc + tf√

m1 = 25 + 8,0√

6,25 = 45 mm (b)

The value of ly using the equation for cases (a) and (b) is clearly larger, thus the valueof lc is the lesser of those above, that is, 45 mm

FCR = 0,9kFEt3w

hw= 0,9 × 4,21 × 210

8,03

68= 5991 kN

λF =√

lytwfyw

FCR=√

45,0 × 8,0 × 3555991 × 103 = 0,146

χF = 0,5λF

= 0,50,146

= 3,42


Leff = χFly = 1,0 × 45 = 45 mm

FRd = Leff twfyw

γM1= 45 × 8,0

3551,0

× 10−3 = 128 kN

The total resistance therefore is 256 kN (from both webs). This exceeds the reactionof 165,2 kN. Thus the length of stiff bearing could be reduced to around 20 mm. Notealso that the end of the bottom member will be sealed by a thin plate welded over itsend to prevent corrosive matter reaching the inside of the tubular section. This willhave a slight stiffening effect.

The results from a fixed-joint analysis are also given in Table 9.1 where it will beobserved that with the exception of the members close to the supports, the axial forceresultants show negligible difference. The exception is member AD which now carriesan axial force of 7 kN and a moment of 2,85 kNm (Mpl,Rd = 35,5 kNm). Thus it maybe concluded the secondary forces induced by joint fixity are negligible, and may beignored. In the fixed-joint analysis the deflections are reduced by around 1–2%.

9.2 NON-TRIANGULATED TRUSSES

Another common form of truss is the Vierendeel girder. This is non-triangulated, eventhough the top and bottom cords may not be parallel. As the loading is carried by sway,

Ch09-H5060.tex 12/7/2007 14: 9 page 348


there are substantial moments at the nodes, although the shears and axial forces ofreasonable size. As a result, the Vierendeel girder is automatically classified as a swayframe and is generally analyzed elastically with the system or effective lengths of themembers determined for the sway case.

The notional horizontal load is applied at the end of the truss in the direction of thetop chord.

EXAMPLE 9.2 Design of a parallel chord Vierendeel girder.

Prepare a design in Grade S355 steel using rectangular hollow section for the girderdetailed in Fig. 9.2.

Loading:

Permanent (kPa): Sheeting 0,22Purlins 0,10Total 0,32

UDL per truss = 0,32 × 8,5 = 2,72 kN/mSelf-weight (estimated) = 2,00 kN/mTotal permanent = 4,72 kN/mor a nodal load of 3,0 × 4,72 = 14,16 kN (internal node)Variable action = 0,6 kPa, or 8,5 × 3,0 × 0,6 = 15,3 kN per nodeTotal factored ultimate load per internal node = 1,35 × 14,16 + 1,5 × 15,3 = 42,07 kN

For this example any load combination involving wind will not be considered. Note,however, where the load cases including wind to be considered then the moments dueto wind would need multiplying by the sway amplification factor.

Determination of notional horizontal loading:

Determine αh from Eq. (8.17):

αh = 2√h

= 2√2

= 1,414

The maximum value allowed for αh is 1,0,

8 bays @ 3 m

B D F H K

J

L N P R

A C E G K M O Q

2,0

m

FIGURE 9.2 Truss geometry for EXAMPLE 9.2

Ch09-H5060.tex 12/7/2007 14: 9 page 349


thus αh = 1,0.

Determine αm from Eq. (8.18):

αm =√

0,5(

1 + 1m

)=√

0,5(

1 + 19

)= 0,745

Determine ϕ from Eq. (8.16):

φ = φ0αhαm = 1,0 × 0,7451

200= 3,725 × 10−3

Total factored vertical loading is 8 × 42,07 = 336,6 kN.

Thus the notional horizontal load is 336,6 × 3,725 × 10−3 = 1,25 kN.

Determination of αcr:

A notional load of 1% of the total factored load is applied at the top chord level. Theload is 0,01 × 336,6 = 3,37 kN.

The actual analysis for both these cases was carried out using a computer package witha 1 kN load, and the results were obtained pro-rata.

The results for the notional horizontal load of 1,25 kN is given in Table 9.3, andthe deflections (absolute and relative) for the notional load of 3,37 kN are given inTable 9.4.

As only a single storey is being considered, then from Table 9.4 the maximum relativedeflection is 0,058 mm. Thus

�

h= 0,058

2000= 29 × 10−6

TABLE 9.3 Internal stress resultants due to the notional horizontal load.

Moment

Axial force Shear force End 1 End 2Member kN kN kNm kNm

AB −0,52 0,093 −0,0845 0,0845CD 0 0,15 −0,1475 0,1475EF 0 0,15 −0,154 0,154GH 0 0,16 −0,155 0,155IJ 0 0,16 −0,155 0,155AC −1,16 −0,05 −0,085 0,072CE −1,00 −0,05 −0,077 0,079EG −0,86 −0,05 −0,075 0,081GI −0,70 −0,05 −0,074 0,083BD 1,16 0,05 0,085 −0,072DF 1,0 −0,05 0,077 −0,079FH 0,85 −0,05 0,075 −0,081HJ 0,70 −0,05 0,074 0,083

Ch09-H5060.tex 12/7/2007 14: 9 page 350


TABLE 9.4 Determination of αcr.

Deflection

Top Bottom Net deflectionMember mm mm mm

AB 0,058 0 0,058CD 0,053 −0,005 0,058EF 0,049 −0,009 0,058GH 0,045 −0,012 0,057IJ 0,042 −0,016 0,058KL 0,040 −0,018 0,058MN 0,038 −0,020 0,058OP 0,037 −0,021 0,058QR 0,037 −0,021 0,058

From Eq. (8.10), the value of αcr is given by

αcr = 0,009�h

= 0,00929 × 10−6 = 310

The minimum value for elastic analysis of αcr is 10, thus second order effects may beignored.

The magnification factor to be applied to moments causing sway (i.e. for this particularexample only the moments due to the notional horizontal loads) is (cl 5.2.2 (5)B, EN1993-1-1)

1

1 − 1αcr

= 1

1 − 1310

= 1,003

The stress resultants due to the vertical loads are given in Table 9.5. The bendingmoment, shear force and axial force diagrams for the total loading are plotted inFig. 9.3(a) to (c), respectively for half the frame

Try a 250 × 150 × 12,5 S355J2H

Member checks:


Web:

The web is subject to compression:

Maximum axial compressive force NEd is 479 kN.

Length of web χw to resist the compressive force is given by

χw = NEd

2twfy

γM0

= 479 × 103

2 × 12,5 3551,0

= 54 mm

Ch09-H5060.tex 12/7/2007 14: 9 page 351


TABLE 9.5 Internal stress resultants due to the factored vertical load.

Moment

Axial force Shear force End 1 End 2Member kN kN kNm kNm

AB −94,53 −116,11 −166,11 116,24CD −21,20 −175,56 −175,56 175,56EF −20,99 −62,52 −124,11 123,90GH −21,04 −62,52 −62,52 62,52IJ −21,04 0 0 0AC −116,11 73,50 115,95 104,59CE −291,67 52,63 −70,93 86,96EG −415,78 31,55 −35,04 57,51GI −478,29 10,52 −5,01 26,55BD 116,11 73,75 116,24 −104,93DF 291,67 52,55 70,72 −86,92FH 415,78 31,55 37,15 −57,51HJ 478,29 −10,52 5,01 −26,55

+73 +55 +32 +10

+73 +53 +32 +10

−116 −175 −123 −62

−117

118

293

293

416

417

479

479

−

−

+ +

− −

+

−95 −21 −21 −21 −21

(a) AFD (kN)

(b) SFD (kN)

(c) BMD (kNm) contension force

176

116

104

104 87 58

537

175

37

87

71

71175 124

124 63 26

2663

0

5

FIGURE 9.3 Final AF, SF and BM diagrams for EXAMPLE 9.2

Ch09-H5060.tex 12/7/2007 14: 9 page 352


c = h − 2tf − 2r = 250 − 2 × 12,5 − 2 × 12,5 = 200 mm

αc = 0,5c − 0,5ax = 100 + 27 = 127 mm

α = 127200

= 0,635

ctw

= 20012,5

= 16

Class 1 limit for α > 0,5:

ctw

= 396

√235fy

13α − 1= 396

√235355

13 × 0,635 − 1= 44,4

Web is Class 1.

Flange:

c = b − 2tf − 2r = 150 − 2 × 12,5 − 2 × 12,5 = 100 mmctf

= 10012,5

= 8

Limit for Class 1 (Table 5.2 Sheet 1):

ctf

= 72

√235fy

= 72

√235355

= 58,6

Flange is Class 1.

The section is therefore Class 1.

Shear:

The shear area Av for a load parallel to the depth is given in cl 6.2.6 (EN 1993-1-1) as

Av = Ah

h + b= 9300

250250 + 150

= 5813 mm2

Vpl,Rd = 1√3

Avfy

γM0= 1√

35813

3551,0

× 10−3 = 1191 kN

The maximum applied shear force is 175,4 kN. Thus the section is satisfactory andthere is no moment capacity reduction due to shear (VEd/Vpl,Rd = 0,15).

Members BD, DF, FH and HJ need checking under combined tension and bending.This check reduces to determining whether the reduced flexural capacity due to axialload exceeds the applied moment.

For box sections:

MN,y,Rd = 1 − n1 − 0,5aw

Mpl,y,Rd

aw = A − 2btA

= 9300 − 2 × 150 × 12,59300

= 0,597

Ch09-H5060.tex 12/7/2007 14: 9 page 353


Limiting maximum value of aw = 0,5, therefore aw = 0,5.

1 − 0,5aw = 1 − 0,5 × 0,5 = 0,75

Mpl,y,Rd = Wpl,yfy

γM0= 751000

3551,0

× 10−6 = 267 kNm

Npl,Rd = Afy

γM0= 9300

3551,0

× 10−3 = 3302 kN

n = NEd

Npl,Rd

The checks are carried out in Table 9.6, where it should be noted that in no case is theplastic moment capacity reduced, and that the values quoted for My,Ed are the largerabsolute values within the member.

Members AB, CD, DF, GH, AC, CE, EG, GI need checking under combined bendingand compressive axial force.

The check for reduction in moment capacity for members AC to GI is exactly the sameas for BD to HJ. Members AB and CD carry lower axial forces and will not thereforebe critical for moment capacity reductions.

Lateral torsional buckling:

The system length may be taken as the worst case of member length, that is, 3 m.Lateral torsional buckling is checked using Section 5.1.8.1.

Determine φb from Eq. (5.63):

φb =

√√√√W 2pl,y

[1 − Iz

Iy

][1 − GIt

EIy

]AIt

=

√√√√(751 × 103)2 [1 − 3310

7518

][1 − 81×7317

210×7518

]9300 × 7317 × 104 = 0,538

The moment gradient factor C1 should be determined for the greatest ratio of endmoments which occurs in member GI.

ψ = −5,0826,63

= −0,191

TABLE 9.6 Member capacity checks for members in tension.

NEd MN,y,Rd Mpl,y,Rd My,EdMember kN n kNm kNm kNm

BD 117,3 0,036 343 267 116,3DF 292,7 0,089 324 267 87,0FH 416,6 0,126 311 267 57,6HJ 479,0 0,145 304 267 26,6

Ch09-H5060.tex 12/7/2007 14: 9 page 354


Determine C1 from Eq. (5.40):

1C1

= 0,6 + 0,4ψ = 0,6 + 0,4(−0,191) = 0,524 > 0,4

Thus C1 = 1,908

The slenderness λ is given by

λ = Liz

= 300059,7

= 50,3

Determine the lateral torsional buckling slenderness ratio λLT from Eq. (5.62):

λLT = 1

C1/21

[π

√EG

]1/2

(φbλ)1/2 = 1√1,908

[π

√21081

]1/2

(0,538 × 50,3)1/2 = 8,47

Determine the normalized slenderness ratio λLT from Eq. (5.60):

λLT = λLT

π

√Efy

= 8,47

π

√210×103

355

= 0,111

As λLT < 0,4, lateral torsional buckling cannot occur.

Check the critical length for lateral torsional buckling lcrit from Eq. (5.64a):

lcrit = 113400 (h − t)fy

(b−th−t

)2

1 + 3 b−th−t

√√√√√3 + b−th−t

1 + b−th−t

= 113400 (250 − 12,5)355

(150−12,5250−12,5

)2

1 + 3 150−12,5250−12,5

√√√√√3 + 150−12,5250−12,5

1 + 150−12,5250−12,5

= 14000 mm

This is well in excess of the system length of 3 m. This also confirms lateral torsionalbuckling will not occur.

Check Eq. (5.64b) from Kaim (2006),

λz,lim = 25hb

√235fy

λz,lim = λz,limλ1 = 25hb

√235fy

93,9

√235fy

= 55167hb fy

= 55167250150 355

= 93,2

L = λz,limiz = 93,2 × 59,7 = 5560 mm

The actual length of 3 m is below the critical value, therefore lateral torsional bucklingwill not occur.

Thus as lateral torsional buckling is not critical Table B.1 can be used to determine thevalues of kyy and kzy.

Ch09-H5060.tex 12/7/2007 14: 9 page 355


TABLE 9.7 Member capacity checks for compression members.

Member

Calculation AC CE EG GI

K C 0,3333 0,3333 0,3333 0,3333K 1 0 0,3333 0,3333 0,3333K 11 0 0 0 0K 12 0,5 0,5 0,5 0,5k1 0,4 0,571 0,571 0,571K 2 0,3333 0,3333 0,3333 0,3333K 21 0 0 0 0K 22 0,5 0,5 0,5 0,5k2 0,571 0,571 0,571 0,571lcr,y/L (Eq. (8.22)) 1,470 1,612 1,612 1,612lcr,y/L (Eq. (8.23)) 1,418 1,571 1,571 1,571lcr,y/L (Eq. (8.24)) 1,473 1,624 1,624 1,624MEd,max 116,03 87,04 57,59 26,63 kNmMEd,min −104,66 −71,01 −35,12 −5,08 kNmN Ed 117,27 292,68 416,63 478,99 kNN cr,y 7979 6565 6565 6565 kNλy 0,643 0,709 0,709 0,709χy 0,873 0,843 0,843 0,843N cr,z 7623 7623 7623 7623 kNλz 0,658 0,658 0,658 0,658χz 0,867 0,867 0,867 0,867kyy 0,407 0,421 0,430 0,570kzy 0,244 0,253 0,258 0,342NEd

χyAfy0,041 0,105 0,150 0,172

kyyMEd,max

Wpl,yfy0,177 0,138 0,093 0,057

Total 0,218 0,243 0,243 0,229NEd

χzAfy0,041 0,102 0,146 0,168

kzyMEd,max

Wpl,yfy0,106 0,083 0,056 0,034

Total 0,147 0,185 0,201 0,202

The calculations for member capacity are carried out in Table 9.7 where it will be seenthat the member capacities are overgenerous. The 2 m verticals (AB, CD, EF and GH)have not been checked as they are shorter than the top chord compression members(therefore have a higher buckling load) and carry lower axial forces.

The strut buckling lengths have been determined assuming the frame can sway. Thisis conservative. The buckling length co-efficients have been determined using Eqs(8.22)–(8.24) for comparison. As will be noted in Table 9.7 There is little differencebetween the results. In this example those from Eq. (8.24) have been used for bucklingin-plane. For out-of-plane buckling, the actual length has been used, although this is

Ch09-H5060.tex 12/7/2007 14: 9 page 356


conservative. To determine the effective lengths, the EI value of the section has beentaken as unity as all the members are fabricated from the same section. Thus EI/Lreduces to 1/L. The axial forces although compression are given as positive, as is thelarger end moment. The smaller one is given the appropriate sign.

Deflection:

Permanent load deflection: 22,4 mm (span/1070)

Variable load deflection: 24,2 mm (span/992)

Total deflection: 44,6 mm (span/538)

These are satisfactory.

Web check (Section 4.8):

The web needs checking at B with the capacity calculated for a single web and thendoubled.

Reaction = 168,3 kN, moment = 116 kNm.


Determine m1:

m1 = fyfbf

fywtw= bf

tw= 75

12,5= 6,0


As ss and c have been assumed to be zero, then lc = 0, then the least value of ly isgiven by

ly = tf

√m1

2= 12,5

√6,02

= 21,7 mm

The depth of the web hw has been taken as

hw = d − 2t − 2r = 250 − 2 × 12,5 − 2 × 12,5 = 200 mm

FCR = 0,9kFEt3w

hw= 0,9 × 2 × 210

12,53

200= 3691 kN

λF =√

lytwfyw

FCR=√

21,7 × 12,5 × 3553691 × 103 = 0,162

As λF < 0,5, m2 = 0.

χF = 0,5λF

= 0,50,162

= 3,1

Ch09-H5060.tex 12/7/2007 14: 9 page 357



Leff = χFly = 1,0 × 21,7 = 21,7 mm

FRd = Leff twfyw

γM1= 21,7 × 12,5

3551,0

× 10−3 = 96,3 kN

The total for both webs is 192,6 kN.

This exceeds the reaction of 168,3 kN.

η2 = FEd

Leff twfyw

γM1

= FEd

FRd= 168,3

192,6= 0,874

η2 = <1,0, therefore the web resistance at A is satisfactory without a stiff bearing.

However, an interaction equation needs checking owing to the co-existence of shearand bending moment:

η2 + 0,8η1 ≤ 1,4


η1 = MEd

Wplfy

γM1

= 116 × 106

751 × 103 3551,0

= 0,435

η2 + 0,8η1 = 0,874 + 0,8 × 0,435 = 1,22 ≤ 1,4


Check the self-weight:

Member lengths (m): Top chord: 24Bottom chord: 24Verticals: 18

Total length = 66 m, mass/unit run = 73,0 kg/m, total mass = 4,82 ton, or very slightlyover the estimated value of 2,0 kN/m. The very slight underestimate will not be criticalas all the members are well overdesigned.

REFERENCES

Kaim, P. (2006). Buckling of members with rectangular hollow sections, In Tubular structures XI (edsJ.A. Packer and S. Willibald.) Taylor and Francis, 443–449.

EN 1991-1-1. Eurocode 1: Actions on structures – Part 1–1: General loads – Densities, self weight,imposed loads for buildings. British Standards Institution/Comité Européen de Normalization.

EN 1993-1-1. Eurocode 3: Design of steel structures – Part 1–1: General rules and rules for buildings.British Standards Institution/Comité Européen de Normalization.

Ch10-H5060.tex 12/7/2007 15: 0 page 358

C h a p t e r 10 / CompositeConstruction

As mentioned in the introduction to Chapter 8, composite construction for slabs andbeams has now become widespread, although some use is still made of pre-cast slabswhich whether supported on the top flange of the beam grid or on shelf angles do notact compositely with the supporting beams (although they will provide lateral torsionalrestraint to the beam). To ensure composite action between the beam and the concreteslab, the two must have adequate shear coupling. In general this is achieved by through-deck stud welding.

A further use of composite construction is concrete filled rolled hollow sectioncolumns. This has the effect of both increasing the load carrying capacity and alsothe fire resistance as the concrete core provides a heat sink, and enables load to betransferred from the steel outer to the core. Construction is fast as the steel sectionitself acts as formwork for the concrete.

Due to developments in alternative, lighter and less time consuming methods of fireprotection, concrete for encasement of steel sections, whether beams or columns, isnow little used in the UK, even though one of the drawbacks, namely extended con-struction time and the need for formwork, can be countered by pre-casting the concreteencasement off-site. There will still exist the problems due to the large additional self-weight. Also earlier design methods did not traditionally make full use of the concretein determining the load carrying capacity (this has changed in EN 1994-1-1).

10.1 COMPOSITE SLABS

A composite slab comprises profile sheet steel decking which acts both as permanentshuttering to the slab and as tension reinforcement for the sagging moments in theslab. There are essentially two basic patterns for profile sheet steel decking; on opentrapezoidal section (Fig. 10.1(a)), and a re-entrant trapezoidal section (Fig. 10.1(b)).There will be variations from these basic shapes dependant upon the specificmanufacturer.

The re-entrant profile has a slight advantage in that proprietary wedge fixings placed inthe re-entrant slot can be used to support suspended ceilings and lightweight services.

Ch10-H5060.tex 12/7/2007 15: 0 page 359


(a) Open trapezoidal section

(b) Re-entrant trapezoidal section

FIGURE 10.1 Types of decking.

The design of such decking when used compositely is considered under two headings,namely those of the decking acting as formwork and the composite slab (Bunn andHeywood, 2004).

10.2 DESIGN OF DECKING

10.2.1 Actions

The actions to be applied to the decking are:

• the weight of the concrete and the self-weight of the deck,• construction and storage loads,• any additional load due to ponding of the deck.

The construction loads are to be taken as 1,5 kPa over an area of 3 m by 3 m (or the spanif less) placed such as to cause the maximum value of the load effect being considered(flexure, shear or deflection) and 0,75 kPa over the remainder of the deck (EN 1991-1-6, cl 4.11.2). These loads are additional to the weight of the concrete. The unit weightof wet concrete should be taken as 1 kN/m3 greater than the dry unit weight (TableA.1 of EN 1991-1-1).

Where the deck is continuous over the support beams, the internal forces shouldbe determined using an elastic analysis with constant values of section properties.Although the properties of the steel sheeting may be calculated from first principles,the calculations are tedious and the information required is generally tabulated by themanufacturer.

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360 • Chapter 10 / Composite Construction

10.2.2 Deflection

If the deflection under self-weight and the wet concrete is less than the least of span/180then any additional effects due to ponding can be ignored (cl. 6.6(2)). If the centraldeflection δ is less than one-tenth of the slab thickness, ponding may be ignored. If thislimit is exceeded then ponding can be allowed for by increasing the nominal thicknessof the concrete by 0,7δ over the whole span.

10.2.3 Design of Composite Slab (EN 1994-1-1)

The analysis may be carried out for ultimate limit state by any recognized method,including simple plastic analysis (cl 9.4.2). If elastic analysis based on uncracked sectionproperties is used, then up to 30% redistribution is allowed. However, it must be notedthat plastic analysis or elastic analysis with a large amount of redistribution will reducethe reinforcement in the top face which may then give problems with achieving anadequate fire performance for other than low fire endurance periods. Any top steel isincreasing mobilized during the course of a fire as the decking and the bottom steel,if provided, loses strength. This situation will be most critical in end spans where thesagging moment can only redistribute to one support during the fire. The designermust effectively choose either to allow redistribution under ambient ultimate loaddesign or in the fire design. Information from tests given are likely to have been basedon an elastic distribution of design moments.

Three basic checks are required:

(1) Flexural failureIt is generally sufficient to design the slab as simply supported between the supportbeam system, although advantage can be taken of continuity. Where the slab isdesigned as simply supported then reinforcement having a cross-sectional area0,2% of the gross concrete section must be provided if the slab is constructed asunpropped, and 0,4% if propped (cl 9.8.1 (2)).

(2) Shear bond failureMost profile decking is manufactured with indents or small shear keys to gener-ate bond to give strain compatibility between the concrete and the steel sheet.The parameters required to check this are determined from tests, and are madeavailable by manufacturers for a specific deck. It should be noted that these param-eters are not defined in the same manner as those for BS 5950 Part 4, and thusit is important that current manufacturer’s information is available (Evans andWright, 1988; Johnson, 2004; Johnson and Anderson, 2004).

The position of the regression line in the two standards is not compatible owingto values of confidence limits used. Equally the parameters derived from BritishStandard tests are concrete strength independent, whereas the intercept definedin the test described in EN 1994-1-1 is not. It would appear that the slopes (mvalues) are similar, but the intercept (k values) need multiplying by approximately1/(0,8fcu)1/2 (= 1,12

√fck) to convert from BS values to EN values with allowance

Ch10-H5060.tex 12/7/2007 15: 0 page 361


for conversion of cube strengths to cylinder strengths (Johnson and Anderson,2004).

(3) Vertical shearThis is rarely critical except in short spans carrying high local loading. Also wherelocal high loads exist punching shear will need checking.

10.2.4 Flexural Design (cl 9.7.2)

10.2.4.1 Sagging

The flexural capacity is determined using a rectangular stress block with a compressivestress in the concrete of 0,85fck/γc. The moment capacity of the slab is then given by

MRd = Np

(dp − x

2

)(10.1)

where Np is the force in the profile decking given by Apfyp/γap where Ap is the area of theprofile decking, fyp is the characteristic yield strength of the decking, γap is the partialsafety factor (= 1,0), dp is the depth to the centroid of the profile decking measuredfrom the extreme compression fibre, and x the depth to the neutral axis is given by

x = Np

0,85b fckγc

(10.2)

where b is the width of the section.

10.2.4.2 Hogging

The concrete deck should be treated as normal reinforced concrete, although it is pos-sible to take the decking can be taken into account if it is continuous over the support.

10.2.5 Shear Bond (Longitudinal Shear) (cl 9.7.3)

The shear bond strength Vl,Rd should be taken as

Vl,Rd = bdp

γvs

(mAp

bLs+ k)

(10.3)

where m and k are empirical design factors obtained from tests, Ls is the shear spanand in this case γvs is taken as 1,25.

The shear span Ls is taken as L/4 for a UDL, for two equal symmetric loads, the distancefrom the support to the nearest load. For other cases Ls can be determined approxi-mately by dividing the maximum moment by the greater vertical shear force adjacentto the support for the span being considered. For internal spans of a continuous slab,Ls may be taken as 0,8 L, and fore external spans 0,9 L where L is the span.

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10.2.5.1 Longitudinal Shear for Slabs with End Anchorage(cl 9.7.4 (3))

The steel sheet must be anchored at the end span by shear studs designed to resistthe tensile force in the steel decking at ultimate limit state. The strength of the studshould be the lesser of its design shear capacity (cl 6.6.4.2, Section 10.3.5) or the bearingresistance Ppb,Rd is given by

Ppb,Rd = kφdd0tfyp

γap(10.4)

where t is the thickness of the sheeting, dd0 is the diameter of the weld collar (whichmay be taken as 1,1 times the stud diameter), kϕ is a factor allowing for the influenceof the deck geometry on the shear strength given by

kφ = 1 + add0

≤ 4,0 (10.5)

where a is the distance from the centre of the stud to the end of the deck which is tobe not less than 1,5dd0.

10.2.6 Flexural Shear Capacity (cl 9.7.5)

The flexural shear capacity Vv,Rd in accordance with EN 1992-1-1 is given by

Vv,Rd = CRd,ck(100ρ1fck

) 13 b0dp (10.6)

where b0 is the mean width of the ribs, k is a shear factor (= 1 + √(200/dp) ≤ 2), ρ1

is the steel ratio (= Ap/b0dp < 0,02), and CRd,c is given by 0,18/γc for normal weightconcrete and 0,15/γc for lightweight concrete (γc = 1,25).

10.2.7 Punching Shear

The punching shear resistance Vp,Rd is given by

Vp,Rd = CRd,ck(100ρ1fck

) 13 cpdp (10.7)

where cp is the punching shear perimeter defined in Fig. 10.2 and hc is the depth ofthe concrete above the ribs.

10.2.8 Deflection (cl 9.8.2)

Deflection checks may be omitted if the span depth ratio does not exceed the appro-priate limit for lightly loaded members (EN 1992-1-1, Table 7.4), and for externalspans provided the load required to produce an end slip between the decking and theconcrete of 0,5 mm exceeds 1,2 times the required service load on the slab.

EXAMPLE 10.1 Composite slab design

A floor system for an office block is to be designed using composite construction. Thebeam layout is given in Fig. 10.3.

Ch10-H5060.tex 12/7/2007 15: 0 page 363


dp

dp

dp

hc

hc

hc

Plan

Loaded area

Critical shear perimeter (radii of hc)

Section

FIGURE 10.2 Criticalshear perimeter. (fromEN 1994-1-1)

4321

A

B

C

D

C1

A1

5 6 7

4000 4000 4000 4000 4000 4000

2500

2500

2500

2000

2500

Flo

or s

pan

FIGURE 10.3 Grid layout for composite deck design.

The profile decking chosen is Richard Lees Holorib, 0,9 mm gauge. The slab is 105 mmthick with lightweight concrete Grade LC25/30. As the usual available length of sheet-ing is 12,5 m and the total span is only 12 m, then the sheeting will be continuous andthe design can consider the beneficial effects of this with two spans of 2,5 m either sideof a central span of 2 m.

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a) Design of deck as shuttering.

To facilitate the design of the sheeting and the composite deck, a series of unit loadUDL cases were analysed with each span loaded. The resultant BMDs and SFDs aregiven in Fig. 10.4.

0,4210,030

0,0080,123

0,585

1,082

0,218 0,009

0,0030,0471,418

BMD

SFD

(b) Span A1 B loaded

(c) Span CB loaded

(d) Beam dimensions

(a) Span A1 loaded

BMD

SFD

0,1910,191

0,0480,309

1,000

1,000

0,096

0,096 0,019

0,019

0,048

BMD

SFD

0,310 0,322 0,019

0,0780,465

1,2450,200

0,039

0,008

0,1241,255

A A1 B C C1 D

2,5 2,5 2,0 2,5 2,5

FIGURE 10.4 BMDs and SFDs for UDL

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In order to asses the maximum moments on the decking, a series of load cases needconsidering:

(1) A constant UDL over the whole system.(2) A UDL of 2,5 m length over the end span (although strictly the construction UDL

should be 3 m long, any extension on to the next interior span will be ignored),(3) A UDL over 2,5 m length symmetric about the first internal support (this is not

quite the worst case, which could be established by using influence lines forcontinuous beams). The BMD and SFD for this case are given in Fig. 10.5.

Assume the dry specific weight of concrete is 19 kN/m3. The wet unit weight is then20 kN/m3.

Take the depth of concrete 105 mm with no allowance for voids.

In determining the moment co-efficients, it is assumed the sagging moment is at mid-span where there is no loading on that span.

UDL due to self-weight of concrete = 0,105 × 20 = 2,1 kPa.

From Fig. 10.4, the sagging moment co-efficients are given in Table 10.1.

Moments due to self-weight of concrete,

Maximum sagging moment = 0,440 × 2,1 = 0,924 kNm/m,

Maximum hogging moment = 0,694 × 2, 1 = 1,458 kNm/m.

The construction load can be taken as 0,75 kPa over the whole system and 0,75 kPaapplied locally either over span AA1 or over the support at A1.

0,292

0,054 0,003

0,013

0,3640,277

0,001

0,006

0,034

0,3551,167

1,145

0,333

BMD

SFD

A

A1 B C C1 D

1,5 1,5

FIGURE 10.5 UDL oversupport A1

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TABLE 10.1 Bending moment co-efficients.

Span Sagging Hoggingloaded AA1 Support A1

AA1 0,585 0,421A1B 0,155 0,310BC −0,024 −0,048CC1 −0,030 0,019C1D 0,004 −0,008Total 0,440 0,694

Uniform UDL:

Sagging: M = 0,440 × 0,75 = 0,330 kNm/m

Hogging: M = 0,694 × 0,75 = 0,521 kNm/m

Local UDL:

Sagging: M = 0,585 × 0,75 = 0,439 kNm/m

Hogging (Fig. 10.5): M = 0,292 × 0,75 = 0,219 kNm/m

Totals due to construction loading:

Sagging: M = 0,330 + 0,439 = 0,769 kNm/m

Hogging: M = 0,521 + 0,219 = 0,740 kNm/m

Due to permanent and construction loading:

MSd+ = 0,924 + 0,769 = 1,693 kNm/m (unfactored)

MSd− = 1,458 + 0,740 = 2,198 kNm/m (unfactored)

From manufacturer’s catalogue:

MRd+ = 6,05 kNm/m (ultimate)

MRd− = 6,23 kNm/m (ultimate)

With a partial safety factor of 1,35 applied to the permanent and construction loading,the capacity of the sheeting to support such loading is more than adequate.

Maximum shear:

This is with the construction load on the end 2,5 m span:

Due to the extra 0,75 kPa:

V = 1,418 × 0,75 = 1,064 kN/m

Due to self-weight and the global 0,75 kPa load:

V = (1,418 + 0,124 − 0,008 + 0,003)(2,1 + 0,75) = 4,38 kN/m

VSd = 1,35 × (4,38 + 1,064) = 7,35 kN/m

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Allowable shear VRd = 95,8 kN/m, satisfactory

Deflection:

Initially consider the end span as simply supported under concrete weight (self-weightof sheeting is negligible).

q = 2,1 kPa; L = 2,5 m; I = 644 × 103 mm4; E = 210 GPa.

The deflection δ on a unit width of slab is given by

δ = 5384

qL4

EI= 5

3842,1 × 2,54

210 × 106 × 644 × 10−9 = 0,008 m

Span180

= 2,5180

= 0,014 mm

Slab thickness/10 = 0,105/10 = 0,011 m, therefore ponding need not be considered.

The deflection is satisfactorily based on simply supported conditions and need not bechecked more accurately.

b) Slab design:

Since the structure of which this slab is a part requires a 90 min fire resistance, redistri-bution will be used in the fire limit state rather than at the ambient ultimate limit state,thus the distribution of moments at ultimate will be taken from an elastic analysis withno redistribution. The concrete self-weight is based on the dry density of 19 kN/m3.

Actions (kPa): Self-weight 2,0

Finishes 2,5

Variable 4,0

Factored permanent load: 1,35(2,0 + 2,5) = 6,08 kPa

Factored variable load: 1,5 × 4,0 = 6,0 kPa.

Moments due to permanent load on all spans:

Sagging: 0,445 × 6,08 = 2,706 kNm/m

Hogging: 0,694 × 6,08 = 4,220 kNm/m

Maximum hogging (Spans AA1, A1B and CC1 loaded):

Moment co-efficient is 0,421 + 0,310 + 0,019 = 0,75

Hogging moment = 0,75 × 6,0 = 4,5 kNm/m.

Maximum sagging (spans AA1, A1B and C1D loaded):

Moment co-efficient is 0,585 + 0,155 + 0,004 = 0,744

Sagging moment = 0,744 × 6,0 = 4,464 kNm/m.

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ULS moments:

Hogging: 4,22 + 4,5 = 8,72 kNm/m

Sagging: 2,706 + 4,464 = 7,17 kNm/m.

Flexural design:

Sagging:

Np = Apfyp

γp= 1597 × 280

1,0= 447200 N/m.

From Eq. (10.2)

x = Np

0,85b fckγc

= 447200

0,85 × 1000 × 251,5

= 31,6 mm

Effective depth, dp = overall depth − depth of NA of sheet = 105 − 16,7 = 88,3 mm

From Eq. (10.1)

MRd = Np

(dp − x

2

)= 447200

(88,3 − 31,6

2

)= 32,4 kNm/m > 7,17 kNm/m

Hogging:

Design as reinforced concrete ignoring sheeting:

The depth allowing for the trapezoidal indents is given by

105 − 0,5 × 51(38 + 12) = 96 mm.

Assume 25 mm cover and 10 mm bars, d = 96 − 25 − 5 = 66 mm.

Assuming αcc = 0,85, then using Eq. (6.22) in Martin and Purkiss (2006),

MSd

bd2fck= 8,72 × 106

1000 × 662 × 25= 0,080

Asfyk

bdfck= 0,652 −

√0,425 − 1,5

MSd

bd2fck= 0,652 −

√0,452 − 1,5 × 0,080 = 0,0997

As = 0,0997 × 1000 × 66 × 25500

= 329 mm2/m.

Fix B385 mesh [385 mm2/m]

Flexural shear:

Maximum shear is at A1 on span AA1:

Co-efficient for complete UDL is

1,418 + 0,124 − 0,019 + 0,019 − 0,008 + 0,002 = 1,537

Shear due to permanent loading = 1,537 × 6,08 = 9,345 kN/m

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Co-efficient for UDL on AB and BC is 1,418 + 0,124 = 1,542

Shear due to variable loading = 1,542 × 6,0 = 9,252 kN/m.

Total shear = 9,345 + 9,252 = 18,6 kN/m.

b0 is the distance between rib centres, that is, 150 mm.

Applied shear per rib = 18,6(150/1000) = 2,79 kN/m

Using Eq. (10.6) to calculate Vv,Rd,

ρ1 = 3851000 × 66

= 5,83 × 10−3

k = 1 +√(

200d

)= 1 +

√(20066

)= 2,74 > 2,0

Vv,Rd = CRd,ck(100ρ1fck

) 13 b0d = 0,15

1,25(2 × 150 × 66) 3

√100 × 5,83 × 10−3 × 25

= 5,8 kN/rib

This is greater than the applied shear.

Shear bond:

For an end span Ls = 0,9 L = 0,9 × 2,5 = 2,25 m, as the loading is a UDL Ls/4 is used,that is, 2250/4 = 562.5 mm.

The values of m (=166,6) and k (=0,005) were from British standard tests, thus whilstthe value of m remains unchanged the value of k has to be amended.

k = 0,005 × 1,12 × f 1/2ck = 0,005 × 1,12 × 251/2 = 0,028

Use Eq. (10.3) to calculate Vl,Rd:

Vl,Rd = bdp

γvs

(mAp

bLs+ k)

= 1000 × 88,31, 25

(166,6 × 15071000 × 625

+ 0,028)

= 32,0 kN/m > VSd

Deflection:

Initially assume there will be no problems with end slip.

From Table 7.4 EN 1992-1-1, basic span depth ratio for the end span of a continuousbeam (or slab) is 26 (lightly stressed).

Actual spaneffective depth ratio

= 2500105 − 16,7

= 28,3

This is higher than the allowable of 26, but since the mid-span capacity is muchhigher than the applied moment, the section will be relatively uncracked thus reducingdeflections.

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End anchorage:

Use 20 mm shear studs.

Tensile force required = Np = 434000 N/m.

Bearing resistance check Eq. (10.4):

For a single shear stud, take a at its minimum distance, that is, 2d0, thus from Eq. (10.5),

kφ = 1 + add0

= 1 + 2 × dd0

dd0= 3 < 4

From Eq. (10.4), the resistance per stud is

Ppb,Rd = kφda0tfyp

γap= 3 × (1,1 × 20) × 0,9 × 280

1,0= 16632 N

No. of studs = Ncf /Ppb,Rd = 447200/16632 = 26,9 studs.

No. of troughs per metre width = 1000/150 = 6,67.

No. of studs per trough = 26,9/6,67 = 4,03. Use 4 per trough.

10.3 COMPOSITE BEAMS

Only the common case of rolled sections with profile steel sheet decking runningeither parallel or transversely to the steel beam will be considered. Through-deckwelded shear studs only will be treated. Continuous beam systems will not themselvesbe considered although both sagging and hogging flexural resistance for compositebeams will be described. For other configurations of floor systems, reference should bemade to the appropriate monograph produced by the Steel Construction Institute (e.g.Mullett, 1992; Mullett and Lawson, 1992; Lawson, et al., 1997; Mullett, 1997).

Although lateral torsional buckling in the sagging zone cannot occur owing to therestraint offered by the continuous composite concrete deck, it is possible for lateralbuckling to occur in the hogging zone at the support. Reference should be madeto Johnson (2004) or Johnson and Anderson (2004) for the background to cl 4.6.2(4). Hogging buckling will not be considered further as it is generally only critical incomposite bridge beams, other than to comment that the likelihood of this is reducedby cross-bracing between adjacent beams at supports.

Although traditionally beams have been considered to be fully connected at the con-crete deck to beam interface, that is, there is no slip, in practice the degree of shearconnection is not full and that some relative slip will occur. This small amount of slipwill not significantly affect either the beam deflection or its ultimate load behaviour(Yam and Chapman, 1968). The acceptance of slip has led to the concept that shearstuds operate at 80% of their design capacity at full interaction. It is possible to reducethis value further to a level of only partial interaction, in which case both the deflec-tion and ultimate load capacity are affected (Johnson and May, 1975). Full interaction

Ch10-H5060.tex 12/7/2007 15: 0 page 371


may require excessive transverse reinforcement for shear and also large numbers ofconnectors giving construction problems.

10.3.1 Section Classification

The section classification limits are determined on the same basis to those for plainsteelwork. It is suggested, however, that the sections used are limited to Class 1, or, ifnecessary, Class 2. If a compression flange is restrained from buckling by attachmentto a concrete flange by shear connectors, it may be assumed to be Class 1 (cl 5.5.2 (1)).

The web slenderness ratio must be checked with allowance being made for the factthat the neutral axis is not at the mid-height of the web. For Class 1 sections the d/twratio should not exceed 396ε/(13α − 1) when α > 0,5 or 36ε/α when α < 0,5. For Class 2sections the constants are 456 and 41,5. The ratio α is the normalized depth of thecompression zone of the web. Where partial interaction is used, the web classificationmust be carried out using full interaction.

10.3.2 Design Criteria

In addition to the expected checks for flexure, vertical shear (carried by the web only),web capacity under in plane forces and deflection, the transverse shear and the shearconnector capacity also need checking.

10.3.3 Flexural Design

For stresses during construction the difference between propped and unproppedmethods need to be noted.

• Propped constructionThe load due to the concrete is not transferred to the composite section until theconcrete has hardened. However, additional forces due to the release of the propsmay also need considering.

• Unpropped constructionThe effect of the wet concrete is taken on the steel beam alone before the onset ofcomposite action.

In both cases the effects of finishes and variable loads are taken on the compositesection.

It should be noted that in unpropped construction the steel beam may still need check-ing for lateral torsional buckling during the construction stage. Where the profiledecking runs parallel to the span of the beam, the sheeting does not provide fullrestraint. In the case of decking perpendicular to the span, full restraint is likely to beavailable (Lawson and Nethercot, 1985).

At ultimate limit state, it is not relevant whether the construction was propped orunpropped.

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Owing to the relatively high flange widths of composite beams, the whole flange cannotbe taken in calculating the moment capacity of the beam owing to shear lag. Theeffective width either side of the web centre line should be considered separately. Thepartial effective width be,i should be taken as Le/8, where Le is the span for simplysupported beams. The total effective width beff is then given as 2be,i + b0, where b0 isthe transverse distance between the shear connectors. In practice the contribution ofb0 may be neglected. The total effective width beff for continuous beams, see cl 5.4.1(EN 1994-1-1). The partial effective width should not exceed either the distance tothe free edge or half the spacing to the adjacent beam centre line.

The following assumptions are made when determining the flexural capacity at ULS:

• The concrete in between the ribs is ignored, and the remainder is stressed with auniform strength of 0,85fck/γc.

• The profile decking may be included if it is in tension and is then stressed to fy/γap,otherwise it is ignored.

• Reinforcement is stressed to fsk/γs where fsk is the characteristic strength of thereinforcement (compression reinforcement may be ignored).

• The steel beam is stressed to a uniform stress of fy/γa.

10.3.3.1 Sagging

All symbols are defined in Fig. 10.6(a).

The force in the concrete Nc is given by

Nc = 0,85fck

γcbeff(hf − hp

)(10.8)

The force in the steel section Na is given by

Na = Aafyγa

(10.9)

where Aa is the cross-sectional area of the steel beam.

There are three possibilities for the position of the plastic neutral axis, in the concreteslab, the top flange of the steel beam or the web of the steel beam.

If the neutral axis lies in the concrete slab, then Nc > Na, otherwise the neutral axis isin the steel beam.

(a) Neutral axis in concrete slab (Fig. 10.6(b)).Depth of neutral axis in the slab x is given by

x = Na0,85fck

γcbeff

(10.10)

The moment capacity Mpl,Rd is then given by

Mpl,Rd = Na

[h2

+ hf − x2

](10.11)

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Concrete slab

beff

Nc

tw

tf

Aa

bf

h f

h p

0,85 fckgc

h AreaxNeutral

axis

Na

Nc

NafxNeutralaxis

(a) Basic data (b) Neutral axis in slab

Nc

Na

Naw

Naf

x

Neutralaxis h�t f

2

h2

(c) Neutral axis in web

(d) Neutral axis in flange

h�t f�

2x

h �2x

x

FIGURE 10.6 Determination of plastic moment capacity (sagg mg).

(b) Neutral axis in the steel beam (Na > Nc)Define an out-of-balance steel compression force Nac as

Nac = Na − Nc (10.12)

and the capacity of the flange Naf (ignoring fillets) as

Naf = bf tffyγa

(10.13)

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The out-of-balance force is introduced to enable the force in the beam Na to be retainedwith its point of application of h/2 in the moment capacity calculations.

The neutral axis is

• in the web if Nac > 2Naf or• in the flange if Nac < 2Naf .

Case 1: Neutral axis in the web (Fig. 10.6(c))

Depth of web x in compression is given by

x = Nac − Naf

2twfyγa

(10.14)

The compression force in the web Naw is given by

Naw = xtwfyγa

(10.15)

and the design moment of resistance Mpl,Rd is given by

Mpl,Rd = Nc

(h + hp + hs

2

)+ 2Naf

(h − tf

2

)+ 2Naw

(h − tf − x

2

)− Na

h2

(10.16)

Case 2: Neutral axis in the flange (Fig. 10.6(d))

Depth of flange x in compression is given by

x = Nac

2bfyγa

(10.17)

The force in the flange Nafx is given by

Nafx = xbfyγa

(10.18)

and the design moment of resistance Mpl,Rd is given by

Mpl,Rd = Nc

(h + hp + hf

2

)+ 2Nafx

(h − x

2

)− Na

h2

(10.19)

10.3.3.2 Hogging (negative moment)

Conservatively the contribution of the steel profile decking will be ignored.

All symbols are defined in Fig. 10.7(a).

The force in the reinforcement Nr is given by

Nr = Asfsk

γs(10.20)

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(c) Neutral axis in top flange

(b) Neutral axis in web(a) Definition of symbols

Na

Na

Naw

Naf

Nr

Nafx

Nr

t f

t f

tw

bf

As

Fs

As

Neutralaxis

Neutralaxis

x

h

c

h

xtftf

h �

x 2

h 2

h � tf �x2h

�2

ReinforcementAs

Steel beam(area As)

FIGURE 10.7 Calculation of the plastic moment capacity (hogging).

The force in the steel beam Na is given by Eq. (10.9), and the force in a flange Naf isgiven by Eq. (10.13). The neutral axis is in

• the web, if Na > Nr + 2Naf ,• the flange, if Na < Nr + 2Naf .

Case 1: Neutral axis in web (Fig. 10.7(b))

The depth of the web x in tension is given by

x = Na − Nr − 2Naf

2twfyγa

(10.21)

The capacity of the tension part of the web Naw is given by

Naw = xtwfyγa

(10.22)

The moment capacity of the section Mpl,Rd is given by

Mpl,Rd = Nr(h + hs − c

)+ Naf

(h − tf

2

)+ Naw

(h − tf − x

2

)− Na

h2

(10.23)

Case 2: Neutral axis in web (Fig. 10.7(c))

The depth of the flange x in tension is given by

x = Na − Nr

2bfyγa

(10.24)

Ch10-H5060.tex 12/7/2007 15: 0 page 376


The capacity of the tension part of the flange Nafx is given by

Nafx = xbffyγa

(10.25)

The moment capacity of the section Mpl,Rd is given by

Mpl,Rd = Nr(h + hf − c

)+ Nafx

(h − x

2

)− Na

h2

(10.26)

10.3.3.3 Reduction in Plastic Moment Capacity Due to Shear

In a similar fashion to plain steel beams, the moment capacity is reduced when the ver-tical shear VEd exceeds half the plastic shear resistance VRd. A reduced design strength(1 − ρ)fyd to determine the capacity of the steel section is used where ρ is defined as

ρ =(

2VEd

VRd− 1)2

(10.27)

For calculation of plastic shear capacity see Section 10.3.4.

10.3.3.4 Reduction in Sagging Moment Capacity with PartialShear Connection (cl 6.2.1.3)

The section capacity should be calculated using a force in the concrete section of Nc,f

where Nc,f is given by

Nc,f = ηNc (10.28)

where η is the degree of shear connection.

The moment capacity Mpl,Rd is then conservatively given by

MRd = Mpl,a,Rd + Nc

Nc,f

(Mpl,Rd − Mpl,a,Rd

)(10.29)

where Mpl,Rd is the moment capacity of the composite section with full shear connec-tion and Mpl,a,Rd is the plastic moment capacity of the steel section alone.

10.3.3.5 Elastic Capacity

For elastic resistance conventional elastic theory is used with the slab taking its effectivewidth beff and where appropriate account is taken of creep (cl 6.2.1.5). The limitingstresses for Mel,Rd are 0,85 fck/γc for concrete, fy/γa for structural steel (Class 1,2 or 3cross-sections) and fsk/γs for reinforcing steel.

Account should be taken of creep and shrinkage in determining the elastic capacity.However, for beams with only one flange composite this may be achieved by using anappropriate modular ratio nL (cl 5.4.2.2.2). The modular ratio nL is defined as

nL = n0 (1 + ψLφt) (10.30)

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where n0 is the short-term modular ratio defined as Ea/Ecm, ϕt is the creep co-efficientdefined as ϕ(t,t0) in EN 1992-1-1, and ψL is creep multiplier depending on the typeof loading (1,1 for permanent loads, 0,55 for shrinkage and 1,5 for prestressing byimposed deformations). For cases where any amplification of internal forces is lessthan 10% due to deformations, the structure is not mainly intended for storage norprestressed by imposed deformations, the effect of creep can be taken into accountfor both short- and long-term loading by using a nominal modular ratio n determinedusing an effective concrete elastic modulus Ec,eff given by 0,5Ecm.

Note, there appears to be no explicit requirement to check serviceability elastic stressesif the plastic moment capacity is used to determine the strength of the beam.

10.3.4 Flexural Shear

The flexural shear capacity Vpl,Rd is calculated exactly as for a normal steel beam.Additionally for an unstiffened and uncased web d/tw should not exceed 69ε, for acased web the constant is 124.

10.3.5 Design of Shear Connectors

The strength of shear connectors is dependant upon both the strength and elasticityof the concrete and the ultimate tensile strength of the connector itself.

Following a large series of tests Olgaard et al. (1971) proposed the following equationfor the strength of shear stud connectors

PRd = kd2√

fcuEc

1,25(10.31)

where d is the diameter of the stud, Ec Young’s Modulus of the concrete, fcu concretecube strength and k is an empirical constant allowing for the height to diameter ratioof the stud. Using the 80% utilization factor proposed by Yam and Chapman (1968),and converting the cube strength to cylinder strength by a factor of 0,8 gives thefollowing strength formula,

PRd = 0,29αd2√

fckEcm

γv(10.32)

where γv takes a value of 1,25 and

for 3 < hsc/d < 4

α = 0,2(

hsc

d+ 1)

(10.33)

for h/d > 4

α = 1,0 (10.34)

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Oehlers and Johnson (1987) carried out further examination of shear stud capacityand proposed an upper bound which was dependant upon the ultimate strength of theshear stud. Their results have been simplified in the code and are given as

PRd = 0,8fuγv

πd2

4(10.35)

where fu is the ultimate tensile strength of the stud material (taken as not greater than500 MPa). For decks with ribs parallel to the beam the strength of the shear studs needsto be reduced by the parameter k1 as the containment by the concrete is incomplete(Mottram and Johnson, 1990),

kl = 0,6b0

hp

(hsc

hp− 1)

≤ 1,0 (10.36)

where hp is the height of the profile and hsc is the overall height of the stud (<hp + 75).

For ribs transverse to the beam the value of fu should be taken as not greater than450 MPa and modified by the parameter kt,

kt = 0,7√nr

b0

hp

(hsc

hp− 1)

(10.37)

where nr is the number of studs in one rib (not greater than 2). For studs with diame-ters not exceeding 20 mm welded through the deck, the maximum values of kt, kt,max

depend on the sheeting thickness and nr. For nr = 1, kt,max = 0,85 for sheeting thick-nesses less than 1,0 mm, and 1,0 for sheeting thicker than 1,0 mm. For nr = 2, kt,max =0,70 for sheeting thicknesses less than 1,0 mm, and 0,8 for sheeting thicker than 1,0 mm.

10.3.6 Biaxial Loading of Shear Connectors

Where the shear connectors are required to produce composite action for both thebeam and the slab, then the following relationship should be satisfied

(Fl

Pl,Rd

)2

+(

Ft

Pt,Rd

)2

≤ 1,0 (10.38)

where Fl is the design longitudinal force caused by composite action in the beam, Ft

is the design transverse force caused by composite action in the slab, Pl,Rd and Pt,Rd

are the corresponding design shear resistances of the stud.

10.3.7 Partial Shear Connection

The code places limits on the amount of partial shear connection that may be used(cl 6.6.1.2).

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For headed studs having a length exceeding 4d with 16 ≤ d ≤ 25 mm, and the steelbeam has equal top and bottom flanges, the minimum value of the ratio η is given by

for Le ≤ 25 m

η ≥ 1 − 355fy

(0,75 − 0,03Le

) ≥ 0,4 (10.39)

for Le ≥ 25 m

η ≥ 1,0 (10.40)

where η is defined by

η = nnf

(10.41)

where n is the actual number of shear connectors, nf is the number required with fullshear connection and Le is the length of the beam subject to sagging bending.

Where the studs have an overall length, after welding of 76 mm, a diameter of 19or 20 mm, are placed centrally in the rib of continuous profiled steel sheeting (withb0/hp ≥ 2 and hp ≤ 60 mm) running perpendicular to a rolled equal flange I or H beamand Nc is calculated from Eq. (10.8), then the following values for η may be used

for Le ≤ 25 m

η ≥ 1 − 355fy

(1,0 − 0,04Le) ≥ 0,4 (10.42)

for Le ≥ 25 m

η ≥ 1,0 (10.43)

10.3.8 Shear Connector Spacing

The spacing should be such that longitudinal shear is adequately transferred andseparation between the slab and beam is prevented.

Ductile shear connectors may be spaced uniformly between critical cross-sections if

• all critical sections are Class 1 or 2,• η satisfies the limits for partial interaction,• the plastic moment resistance of the section does not exceed 2,5 times the plastic

moment resistance of the steel section alone.

10.3.9 Longitudinal Shear Force

This is based on the capacity of the shear studs within the length being considered.

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The longitudinal shear force/per unit run VSd is given by

VSd = nrPRd

sL(10.44)

where PRd is shear capacity of the stud, nr is the number of studs in a cross-sectionand sL is the longitudinal spacing of the studs along the beam.

10.3.10 Transverse Shear

This arises due to the transmission of the longitudinal shear in the beam to the slab.The Eurocode for steel–concrete composite design refers to EN 1992-1-1 cl 6.2.4 anduses a truss analogy to determine the reinforcement required.

The critical planes that need checking are illustrated in Fig. 10.8.

For the surface 2–2 the length of the shear surface is taken as 2 h plus the head diameterfor a single row of studs or 2h + st where st is the transverse spacing of the studs (h isthe height of the stud). The shear stress VSd that needs to be resisted is given by theshear force per unit length (VSd) divided by the length of the shear plane (s).

The amount of reinforcement per unit run Asf /sf is determined from

Asf fyd

sf>

vEdhf

cot θf(10.45)

where θf is the angle of inclination of the concrete truss member such that1,0 ≤ cot θf ≤ 2,0.

For normal weight concrete the shear resistance vEd is given by

vEd = 0,5vfcd (10.46)

where v is given by 0,6 (1 − fck/250).

For lightweight concrete the shear resistance vEd is given by

vEd = 0,5η1v1 fcd (10.47)

where v1 is given by 0,5η1(1 − fck/250), and η1 is given by 0,4 + 0,6ρ/2200 where ρ isthe density of the concrete.

A limit is placed on vEd such that

vEd < vfcd sin θf cos θf (10.48)

It would appear acceptable to take the traditional approach and use an angle of 45◦

in the truss analogy giving cot θf = 1,0 and sin θf = cos θf = 1/√

2 (Johnson, 2004).

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5

5

4

3

3

1

1

2 2

4

Plane1 � 12 � 2

Steel areaAb � At

2Ab

Plane4 � 45 � 5

Steel area

2At

Plane3 � 3

Steel area2At

Lab joint in sheeting

Ab

At

At

At

(a) Solid slab

(b) Composite slab (sheeting perpendicular to span)

(c) Composite slab (sheeting parallel to span)

Sheeting

FIGURE 10.8 Criticaltransverse shear planes.

Where the profile sheet decking is normal to the span and is continuous over the beamthen Eq. (10.45) applied to vertical shear planes may be enhanced by the effect of thedecking to give

Asf fyd

sf+ Apefyp,d >

vEdhf

cot θf(10.49)

where Ape is the effective cross-sectional area of the decking and fyp,d is its designstrength.

Ch10-H5060.tex 12/7/2007 15: 0 page 382


hh s

h p

beff

Concrete slab

Steel beam(second moment of area,Iaarea, Aa, modular ratio n)

FIGURE 10.9 Second moment of area calculation for a composite beam.

10.3.11 Deflection (cl 5.2.2)

For the deflection of the steel member alone the principles of EN 1993-1-1 are applied.

No account need be taken of partial interaction if η ≥ 0,5 or the force in the shearconnector does not exceed PRd at the serviceability limit state, and for ribbed slabstransverse to the beam the rib height does not exceed 80 mm.

The composite second moment of area may be determined as follows (Fig. 10.9)

Ic = nIa + beff(hs − hp

)312

+ nAa

(h2

− x)2

+ beff(hs − hp

) (h + hs + hp

2− x)2

(10.50)

where x is given by

x =nAa + (hs − hp

)beff

(h + hs+hp

2

)nAa + (hs − hp

)beff

(10.51)

10.3.12 Vibration

This is only likely to be critical on lightly loaded, long span beams. It is suggested thatthe following formula is used (Wyatt, 1989)

f = 18√δsw

(10.52)

where f is the frequency (Hz) and δsw is the instantaneous deflection (mm) due to self-weight and permanent loading. The suggested limits for f are 4 Hz for most buildings,3 for car parks and 5 for sports halls (Lawson and Chung, 1994).

10.3.13 Detailing

10.3.13.1 Cover

This should be the greater of 20 mm or the values specified in EN 1992-1-1 (Table 4.4)less 5 mm (cl 6.6.5.2).

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10.3.13.2 Spacing (cl 6.6.5.5)

This should be not less than 22tfε for a solid slab or 15tfε for a ribbed slab where theflange becomes a Class 1 or Class 2 by virtue of restraint due to shear connection withthe edge distance not exceeding 9tfε. The maximum spacing should not exceed thelesser of 800 mm or six times the slab thickness. The edge distance should also notexceed 20 mm.

The overall height of a stud connector should be not less than 3d (where d is the shankdiameter), the spacing in the direction of the shear force should be not less than 5d,and transverse to the shear force 2,5d for solid slabs and 4d for other cases. Unless thestud is directly over the web the diameter of the stud should not exceed 2,5tf (cl 6.6.5.7).

For through-deck welding, the profile sheet steel decking should not exceed 1,25 mmthick if galvanized and 1,5 if not. The studs should extend 2d above the deck afterwelding.

Johnson (2005) expresses concern over the detailing rules (and stud capacities) whenapplied to open trapezoidal decks.

EXAMPLE 10.2 Composite beam (decking transverse to span)

Design a composite beam (ref C1/23 (Fig. 10.3)) in Grade S355 steel to carry thecomposite slab designed in Example 10.1. The concrete is lightweight LC25/30 and adry specific weight of 19 kN/m3. The span is 4 m and carries the two 2,5 m deck spans.

Loading due to wet concrete: 0,105 × 19 = 2,0 kPa

Finishes: 2,5 kPa

Variable loading: 4,0 kPa

The loading co-efficients for the reaction at A1 (from SFDs in Fig. 10.4) are given inTable 10.2 for all spans loaded.

For the variable loading, spans AA1, A1B, CC1 and C1D are loaded. This gives anoverall co-efficient of 3,062

TABLE 10.2 Loading co-efficients for reactions A1.

Span Individual components Total

AA1 1,418 + 0,218 1,636A1B 0,124 + 1,254 1,369BC −0,019 − 0,096 −0,115CC1 0,039 + 0,008 0,047C1D 0,009 + 0,003 0,012Overall total 2,949

Ch10-H5060.tex 12/7/2007 15: 0 page 384


Load per unit run on the beam:

Permanent: 2,949(2,0 + 2,5) = 13,27 kN/m

Variable: 3,062 × 4 = 12,25 kN/m

Ultimate limit state design:

MSd = (1,35 × 13,27 + 1,5 × 12,25) × 42/8 = 72,6 kNm

Effective width of slab, beff (ignoring spacing between connectors as it is likely only asingle row will be needed)

be,i = Le/8 = 4000/8 = 500 mm

beff = 2be,i = 1000 mm.

Try a 203 × 133 × 25 UKB (Grade S355)

Since the flange is restrained by through deck studs, the beam flanges are automaticallyClass 1. The web slenderness satisfies the shear buckling check.

Actual width between beam centre lines = 2500 mm > beff

Dimensions to calculate Mpl,Rd assuming full shear connection are given in Fig. 10.10.

From Eq. (10.8), the force in the concrete flange Nc is given by

Nc = 0,85fck

γcbeff(hf − hp

) = 0,85 × 251,5

1000 × (105 − 51) × 10−3 = 765 kN

From Eq. (10.9) the force in the steel beam Na is given by

Na = Aafyγa

= 32003551,0

× 103 = 1136 kN

From Eq. (10.12) the out-of-balance force Nac is given by

Nac = Na − Nc = 1136 − 765 = 371 kN

133,2

5,7

7,8

7,8

1000

105

51

203,

2

Concrete slab

Steel beam203�133�25 UKB S 375

FIGURE 10.10 Design dimensions for EXAMPLE 10.2

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From Eq. (10.13) the capacity of the flange Naf is given by

Naf = btffyγa

= 133,2 × 7,83551,0

× 10−3 = 369 kN

Since Nac < 2Naf , the neutral axis is in the flange.

From Eq. (10.17) the depth of flange x is given by

x = Nac

2bfyγa

= 371 × 103

2 × 133,2 × 3551,0

= 3,92 mm

From Eq. (10.18) the compression force in the flange Nafx is given by

Nafx = xbfyγa

= 3,92 × 133,23551,0

× 10−3 = 185 kN

From Eq. (10.19) Mpl,Rd is given by

Mpl,Rd = Nc

(h + hp + hf

2

)+ 2Nafx

(h − x

2

)− Na

h2

= 0,765(

203,2 + 105 + 512

)+ 2 × 0,185

(203,2 − 3,78

2

)− 1,136

203,22

= 174,2 kNm

Note, for convenience all the forces have been expressed in MN. Then if the dimensionsare left in mm, the moment is in kNm.

Since the neutral axis for full shear connection is in the flange, the only web check isfor shear buckling (see above).

As Mpl,Rd exceeds MEd by a substantial margin, partial interaction may be used.

For the beam alone the section is Class 2 (which is satisfactory).

Check shear capacity:

Vpl,Rd = 1√3

Avfyγa

= 1√3

12803551,0

× 10−3 = 262 kN

VEd = (1,35 × 13,27 + 1,5 × 12,25) × 42

= 72,6 kN

Since VSd < 0, 5Vpl,Rd, there is no reduction in moment capacity.

Shear connectors:

Use 100 mm long by 19 mm diameter stud connectors.

The load capacity of a single stud, Eq. (10.32):

h/d = 100/19 = 5,3, so from Eq. (10.34), α = 1,0

fck = 25 MPa

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For normal weight concrete,

Ecm = 22(

fck + 810

)1/3

= 22(

25 + 810

)1/3

= 31,5 GPa

Modification factor for lightweight concrete ηE (from cl 11.3.2 of EN 1992-1-1)

ηE =( ρ

2200

)2 =(

19002200

)2

= 0,746

So, Ecm = 0,746 × 31,5 = 23,5 GPa

From Eq. (10.32)

PRd = 0,29αd2√

fckEcm

γv= 0,29 × 1 × 192 ×√25 × 23,5 × 103

1,25× 10−3 = 64,2 kN

Limiting capacity:

For ribs transverse to beam the ultimate connector strength (fu = 450 MPa) needs tobe multiplied by the factor kt.

Determine kt from Eq. (10.37)

In view of the large overdesign on moment capacity, number of connectors per rib willbe 1, that is, nr = 1.

For a dovetail deck b0 is the distance between the top of the dovetails, that isb0 = 150 − 38 = 112 mm

Profile height hp = 55 mm, height of connector, h = 100 mm,

kt = 0,7√nr

b0

hp

(hsc

hp− 1)

= 0,7√1

11255

(10055

− 1)

= 1,17

However the maximum value allowed for kt is 1,0.

Limiting capacity of studs from Eq. (10.35)

PRd = 0,8fuγv

πd2

4= 0,8 × 450

1,25π × 192

4× 10−3 = 81,7 kN

Shear stud capacity is the lower of the two values, that is 64,2 kN.

No. of connectors for full shear connection:

Nc = 765 kN, PRd = 64,2 kN, thus nf for the half beam span is given by

nf = Nc/PRd = 765/64,2 = 11,9

Use 1 connector per rib, thus n for the whole beam is given by span over distancebetween centre lines of ribs = 4000/300 = 13,3.

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The ratio between η between n and nr is given by

η = nnr

= 13,32 × 11,9

= 0,56

The limiting value of η is given by Eq. (10.39) with Le = 4,0 m, thus

η ≥ 1 − 355fy

(0,75 − 0,03Le

) ≥ 0,4 = 1 − 355355

(0,75 − 0,03 × 4

) = 0,37

The limiting value is 0,4, as the actual value is 0,56, it is therefore satisfactory.

The moment capacity MRd is given by Eq. (10.29).

The moment capacity of the steel section Mpl,a,Rd alone is given by

Mpl,a,Rd = 258 ×(

3551,0

)× 10−3 = 91,6 kNm

MRd = Mpl,a,Rd + nc

nc,f

(Mpl,Rd − Mpl,a,Rd

) = 91,6 + 0,56 (174,2 − 91,6)

= 137,9 kNm

This is greater than MEd.

Longitudinal shear:

The only plane requiring checking is 4–4 (or 5–5) of Fig. 10.8

Determine VSd from Eq. (10.44),

VSd = nrPRd

sL= 1,0 × 64,2

0,3= 214 kN/m

The shear stress VSd is determined using the thickness of the concrete above the ribs,that is, 105 − 51 = 54 mm. Also the shear may be equally divided between the twoshear planes, thus VEd is given by

vEd = 0,5 × 21454

= 1,98 MPa

As the sheeting is continuous across the beam with the ribs running normal to thebeam, the contribution of the sheeting may be mobilized.

Using a 45◦ angle for the truss analogy, cos θf = sin θf = 1/√

2 and cot θf = 1,0.

For lightweight concrete vEd is given by Eq. (10.47).

Determine v1:

The reduction factor for lightweight concrete η1 is given by

η1 = 0,4 + 0,6ρ

2200= 0,4 + 0,6

19002200

= 0,918

v1 = 0,5η1

(1 − fck

250

)= 0,5 × 0,918

(1 − 25

250

)= 0,413

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VEd = 0, 5v1fck

γc= 0,5 × 0,413 × 25

1,5= 3,44 MPa

From Eq. (10.48) the maximum value of VEd is given by

VEd = vfcd sin θr cos θr = 0,6(

1 − 25250

)251,5

1√2

1√2

= 4,5 MPa

Use Eq. (10.49) to determine the requirement for Asfsf as the sheeting runs normalto the span.

Evaluate the right hand side of Eq. (10.49),

vEdhf

cot θf= 3,44 × 55

1,0= 1892,2 N/mm

Evaluate the contribution of the sheeting:

Apefyp,d = 15971000

2801,0

= 447,2 N/mm

The sheeting overprovides the required resistance, therefore only minimum reinforce-ment is necessary.

Deflections and service stresses:

a) Wet concrete:

This is taken on the steel beam alone. The uniformly distributed load due to the wetconcrete q is given by

q = 20 × 0,105 × 2,5 = 5,25 kN/m

Deflection, δconc is given by:

δconc = 5384

qL4

EI= 5

3845,25 × 44

210 × 106 × 2340 × 10−4 = 3,6 × 10−3 m

MEd = qL2

8= 5,25 × 42

8= 10,5 kNm

The numerical value of the stress σconc is given by

σconc = MEd

Wel= 10,5 × 103

230= 46 MPa

b) Variable and permanent loads:

As the ratio n/nf of the number of shear studs provided to that required for fullconnection is 0,57 (and is therefore greater than the critical ratio of 0,5), no accountneed be taken of partial shear connection in determining the deflection.

Assume Ec,eff = Ecm/2 = 23,5/2 = 11,75 GPa

The value of n from cl 5.4.2.2 (11) is Es/Ec,eff (=210/11,75 = 17,9)

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Use Eq. (10.51) to determine the elastic centroidal axis x,

x =nAa + (hs − hp

)beff

(h + hs+hp

2

)nAa + (hs − hp

)beff

=17,9 × 3200 203,2

2 + (105 − 51)

2500(

203,2 + 105+512

)17,9 × 3200 + (105 − 51

)2500

= 227, 7 mm

From Eq. (10.50)

Ic = nIa + beff(hs − hp

)312

+ nAa

(h2

− x)2

+ beff(hs − hp

) (h + hs + hp

2− x)2

= 17,9 × 23,4 × 106 + 2500(105 − 51

)312

+ 17,9 × 3200(

203,22

− 227,7)2

+ 2500(105 − 51

) (203,2 + 105 + 51

2− 227,7

)2

= 1,75 × 109 mm4

IcEc,eff = 1,75 × 109 × 11,75 = 20,45 × 109 kNmm2 = 20,56 × 103 kNm2

Variable load is 13,27 kN/m, and the moment MEd is 26,54 kNm.

Thus the deflection δv is given by

δv = 5384

qL4

EI= 5

38413,27 × 44

20,56 × 103 = 2,15 × 10−3 m

This is equivalent to span/1860 which is acceptable.

Although not strictly necessary, determine the stresses due to the variable load:

Top of the concrete slab σv,top,c:

σv,top,c = −MEd(h + hs − x

)Ic

= −26,54 × 106 (203,2 + 105 − 227,7)

1,75 × 109 = −1,22 MPa

Stress at the soffit of the steel beam, σv,soffit:

σv,soffit = nMEdxIc

= 17,9 × 26,54 × 106 × 227,71,75 × 109 = 61,8 MPa

Stress at the top of the steel beam, σv,top,A:

σv,top,c = −nMEd(h − x)Ic

= −17,9 × 26,54 × 106 (203,2 − 227,7)

1,75 × 109 = 6,65 MPa

Permanent load is 12,07 kN/m, and the moment MEd = 24,14 kNm.

δv = 5384

qL4

EI= 5

38412,07 × 44

20,45 × 103 = 2,0 × 10−3m

Total deflection:

δtotal = δp + δv + δconc = 2,2 + 2,0 + 3,6 = 7,8 mm

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This is equivalent to span/513 which is acceptable.

Although not strictly necessary determine the stresses due to the variable load:

Top of the concrete slab σp,top,c:

σp,top,c = −MEd(h + hs − x

)Ic

= −24,14 × 106 (203,2 + 105 − 227,7)

1,75 × 109

= −1,11 MPa

Stress at the soffit of the steel beam, σp,soffit:

σp,soffit = nMEdxIc

= 17,9 × 24,14 × 106 × 227,71,75 × 109 = 56,2 MPa

Stress at the top of the steel beam, σp,top,A:

σp,top,c = −nMEd(h − x

)Ic

= −17,9 × 24,14 × 106 (203,2 − 227,7)

1,75 × 109 = 6,05 MPa

Final total stresses:

Top of concrete slab: −1,11 − 1,22 = −2,33 MPa

Top of steel beam: −46 + 6,65 + 6,05 = −33,3 MPa

Soffit of steel beam: 46 + 61,8 + 56,2 = 164 MPa

All these stresses are acceptable.

Vibration:

Use ψ2 = 0,3, so quasi-permanent load is 12,07 = 0,3 × 13,27 = 16,05 kN/m

Deflection under this load δsw is given by

δsw = 5384

qL4

EI= 5

38416,05 × 44

20,45 × 103 = 0,0026 m = 2,6 mm

Determine the frequency f from Eq. (10.52)

f = 18√δsw

= 18√2,6

= 11,2 Hz

This is well above the recommended limit of 3 Hz.

If the full variable load is taken, then f = 8,7 Hz (and is still acceptable).

EXAMPLE 10.3 Composite beam design (decking parallel to the span).

Prepare a design in Grade S355 steel for the beam Mark 3AD of Fig. 10.3 where thereare no intermediate columns. The actions on the beam are given in Fig. 10.11. Thecomposite deck is that of EXAMPLE 10.1.

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Due to the long span, the beam is designed as propped to eliminate the high deflectionsunder the permanent loading due to the wet concrete. The resultant shear force andbending moment diagrams for the applied loading are given in Fig. 10.12.

The effective width, either side of the beam centre line, be,i = Le/8 = 12000/8 = 1500 mm.Thus total effective width beff = 2be,i = 3000 mm. (The actual beam spacing is 4 m.)

To determine the capacity of the beam, the effect of the voids in the deck will beignored as these run in the direction of the beam and are small.

Try a 838 × 292 × 194 Grade S355 UKB.

94

70

94

70

214

240

214

240

2,5 2,5 2,5 2,52,0

Permanent (kN)

Variable (kN)

FIGURE 10.11 Loading forEXAMPLE 10.3

2500

232 649 649 232 All loads in kN

2500 2000 2500 2500

2203

881

649

649

881

3825

BMD (kNm)

SFD (kN)

FIGURE 10.12 BM and SF EXAMPLE 10.3

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Moment capacity with full shear connection:

Determine the force in the concrete Nc from Eq. (10.8),

Nc = 0,85fck

γcbeff(hf − hp

) = 0,85 × 251,5

3000 × 105 × 10−6 = 4,463 MN

Determine the force in the steel section Na from Eq. (10.9):

Na = Aafyγa

= 247003551,0

× 10−6 = 8,769 MN

Determine the maximum force in a flange Naf from Eq. (10.13):

Naf = bf tffyγa

= 292,4 × 21,73551,0

× 10−6 = 2,253 MN

Determine the out-of-balance force Nac from Eq. (10.12):

Nac = Na − Nc = 8,769 − 4,463 = 4,306 MN

As Nac < 2Naf , therefore neutral axis lies in the flange.

Determine the position of the neutral axis x from Eq. (10.17):

x = Nac

2bfyγa

= 4,306 × 106

2 × 292,4 3551,0

= 20,74 mm

Determine the compression force Nafx in the flange from Eq. (10.19):

Nafx = xbfyγa

= 20,74 × 292,43551,0

× 10−6 = 2,153 MN

Determine the plastic moment capacity Mpl,Rd from Eq. (10.19):

Mpl,Rd = Nc

(h + hp + hf

2

)+ 2Nafx

(h − x

2

)− Na

h2

= 4,463(

840,7 + 1052

)+ 2 × 2,153

(840,7 − 20,74

2

)− 8,769

840,72

= 3876 kNm

This is greater than MEd (=3825 kNm).

Check the ratio of Mpl,Rd to Mpl,a,Rd (the moment capacity of bare steel section):

Mpl,Rd

Mpl,a,Rd= Mpl,Rd

Wplfyγa

= 3876

7640 3551,0 × 10−3

= 1,43

This is less than the critical value of 2,5.

Since the neutral axis is in the flange, the web check is unnecessary and thus the sectionis Class 1.

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Flexural shear:

Vpl,Rd = 1√3

Avfyγa

= 1√3

131003551,0

× 10−3 = 2685 kN

VEd = 881 kN (<0,5Vpl,Rd), thus there is no reduction of moment capacity.

Shear connector design:

As the margin between Mpl,Rd and MEd is small (Mpl,Rd/MEd = 3876/3817 = 1,015),there will be full shear connection.

Use 100 mm long by 19 mm diameter stud connectors.

Load capacity of a single stud:

h/d = 100/19 = 5,3, so from Eq. (10.34) α = 1,0

With fck = 25 MPa (as in EXAMPLE 10.2), Ecm = 23,5 GPa.

From Eq. (10.32)

PRd = 0,29αd2√

fckEcm

γv= 0,29 × 1 × 192 ×√25 × 23,5 × 103

1,25× 10−3 = 64,2 kN

Determine the reduction factor, kl from Eq. (10.36):

With b0 = 150 − 38 = 112 mm; hp = 55 mm; h = 100 mm,

kl = 0,6b0

hp

(hsc

hp− 1)

= 0,611255

(10055

− 1)

= 1,0

Limiting capacity:

Limiting capacity of studs from Eq. (10.35):

For ribs parallel to beam fu = 500 MPa,

PRd = 0,8fuγv

πd2

4= 0,8 × 500

1,25π × 192

4× 10−3 = 90,8 kN

Shear stud capacity is the lower of the two values, that is, 64,2 kN.

Number of connectors for full shear connection:

n = Nc/PRd = 4,463 × 103/64,2 = 70 per half beam.

Use studs in pairs transversely, so sL = 6000/(70/2) = 171 mm.

Use a spacing of 150 mm.

Transverse Shear:

The only plane requiring checking is 4–4 (or 5–5) of Fig. 10.8

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Determine VSd from Eq. (10.44),

VSd = nrPRd

sL= 2 × 64,2

0,150= 856 kN/m

Assume the sheeting rib is along the centre line of the beam, vEd is determined usingthe total thickness of the concrete. Also the shear may be equally divided between thetwo shear planes, thus vEd is given by vEd = 0,5 × 856/105 = 4,08 MPa.

This must be resisted entirely by reinforcement as the ribs of the sheeting are runningparallel with the beam.

Using a 45◦ angle for the truss analogy, cos θf = sin θf = 1/√

2 and cot θf = 1,0.

As the concrete is the same as EXAMPLE 10.2, vEd = 3,44 MPa, and the maximum valueis again 4,5 MPa.

Use Eq. (10.49) to determine the requirement for Astsf :

Asf

sf>

1fyd

vEdhf

cot θf= 1,15

5004,08 × 105

1,0= 0,985 mm

This can be divided between top and bottom, so Ast/Sr on each face is 0,493 mm2/mm.

Use B503 Mesh [503 mm2/m]

Deflection:

The loading for determining variable and total deflections is given in Fig. 10.11.

Use the formula δ = (WL3/48EI)(3(a/L) − 4(a/L)3)

To simplify calculations determine L3/48EI as a constant for the composite and steelsections, and then determine W (3(a/l) − 4(a/L)3) for each load.

Determination of Ic:

Neglect effect of profiles and take hp = 0, beff = 3000 mm and αe = 17,9 (asEXAMPLE 10.2).

Summarizing results: x = 396,3 mm, Ic = 0,162 × 1012 mm4.

Ec,effIc = 11,75 × 106 × 0,162 = 1,904 × 106 kNm2

L3/48Ec,effIc = 123/48 × 1,904 × 106 = 18,9 × 10−6 m/kN

Loads at A1:

a/L = 2,5/12 = 0,208; 3(a/L) − 4(a/L)3 = 0,589

Total load:

W = 94 + 70 = 164 kN,

δ = 164 × 0,589 × 18,9 × 10−6 = 0,0018 m

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Loads at B:

a/L = 5/12 = 0,417; 3(a/L) − 4(a/L)3 = 0,961

Total load:

W = 214 + 240 = 454 kN,

δ = 454 × 0,961 × 18,9 × 10−6 = 0,0082 m

Final deflection under total loads:

δ = 2(0,0018 + 0,0082) = 0, 0202 m

Span/deflection ratio is 12/0,0202 = 594. This is acceptable.

Vibration:

Total variable deflection = 10,2 × 10−3 m

Total permanent deflection = 9,87 × 10−3 m

Variable deflection due to a value of ψ2 = 0,3:

0,3 × 10,2 × 10−3 = 0,0031 m

δsw = 0,0031 + 0,0102 = 0,0133 m = 13,3 mm

From Eq. (10.52), f = 18/√

13,3 = 4,94 Hz

This is higher than the minimum recommended value of 3 Hz.

If the total variable load is taken then f = 4,0 Hz, which is still acceptable.

10.4 COMPOSITE COLUMNS

These can take a variety of forms but fall essentially into two categories; partiallyor totally encased Universal Columns (or H sections) and filled rolled hollow sec-tions, with or without additional reinforcement. Typical configurations are given inFig. 10.13. This text only considers composite columns which are symmetric aboutboth axes.

The methods given in EC 1994-1-1 only hold if

(a) the steel contribution ratio δ defined as

δ = Aa fyd

Npl,Rd(10.53)

satisfies the limits 0,2 ≤ δ ≤ 0,9. For δ < 0,2 the column should be designed asreinforced concrete, and for δ > 0,9 designed as non-composite steel.

(b) the normalized slenderness ratio λ is less than 2,0.

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Traditional encasedcolumn

Web in filledcolumn

Concrete unfilledhollow section

FIGURE 10.13 Types of compositecolumns.

10.4.1 Axial Compression

The design condition for axial compression is that the design resistance χNpl,Rd shouldexceed the applied load NEd. The buckling co-efficient χ is determined using a non-dimensionalized slenderness ratio λ (defined in Eq. (10.73)) in combination withbuckling curve ‘a’ for concrete filled hollow sections with ρs ≤ 3% or ‘b’ 3% < ρs < 6%(whereρs is the percentage of reinforcement), ‘b’ for partially or fully encased I sectionsbending about the major axis, and ‘c’ for partially or fully encased sections bendingabout the minor axis (the buckling curve designations are those used in EN 1993-1-1).

10.4.2 Uniaxial Bending and Axial Compression

Initially, the resistance of the cross-section is determined using an interaction diagrambetween the axial load resistance and bending moment resistance in a similar fashionto reinforced concrete columns. The diagram is shown schematically in Fig. 10.14,where Npl,Rd is the axial squash capacity (Point A), and Mpl,Rd is the plastic momentcapacity (Point B). Although the interaction diagram which is of a similar shape tothat for reinforced concrete columns is strictly curved, it may be approximated to aseries of straight lines. Point C is established by the application of the moment Mpl,Rd

and a resultant axial capacity of the concrete alone Npm,Rd, Point D by the momentcapacity Mmax,Rd under 0,5Npm,Rd.

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N

A

M

D

B

C

Npl,Rd

Npm,Rd

½Npm,Rd

Npl,Rd Mmav,Rd

FIGURE 10.14Interaction diagram forcomposite columns

The moment capacity μdyMpl,y,Rd (or Mpl,N,Rd) is determined from the interactiondiagram under an axial load of NEd. The member is deemed to have sufficient capacitywhen

MEd

Mpl,N,Rd= MEd

μdMpl,Rd= αM (10.54)

The co-efficient αM takes a value of 0,9 for steel grades of S235 and S355 and 0,8 forgrades S420 and S460. The factor is partially needed to compensate for the assumptionover the depth of the rectangular stress block between EN 1992-1-1 where it is taken asextending over 0,8x but over x in EN 1994-1-1, since the same stress level of 0,85fck/γc

is used in both, and partially to allow for the adverse effect of the higher yield strainon the crushing on the concrete.

Should the bending moment be entirely due to the eccentricity of the axial load, thenμd can be greater than unity.

10.4.3 Biaxial Bending

The procedure is similar to uniaxial bending except that two parameters μdz and μdy

now need to be determined. However, the effect of imperfections needs only to beconsidered for the likely failure axis (usually the minor or zz axis). The column is thensatisfactory if the following conditions are satisfied,

My,Ed

μdyMpl,y,Rd≤ αM,y (10.55)

Mz,Ed

μdzMpl,z,Rd≤ αM,z (10.56)

and

My,Ed

μdyMpl,y,Rd+ Mz,Ed

μdzMpl,z,Rd≤ 1,0 (10.57)

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The co-efficients αM,y and αM,z are to be taken as αM. The increase of the interactionco-efficient to 1 in the combined equation is due to the higher crushing strength of theconcrete under biaxial bending.

10.4.4 Determination of Member Capacities

The formulae for the flexural capacity of concrete filled rectangular tubes (withreinforcing steel ignored) are taken from Johnson and Anderson (2004).

10.4.4.1 Axial Squash Capacity, Npl,Rd [Point A]

This is given by the sum of the individual components due to the steel section, theconcrete and the reinforcement (Fig. 10.15(a)). So in general Npl,Rd is given by

Npl,Rd = Aafyγa

+ Ac0,85fck

γc+ As

fsk

γs(10.58)

where Aa is the area of the steel section and fy is the yield strength, Ac is the concretearea and fck is the characteristic cylinder strength, As is the area of the reinforcementand fsk is the characteristic strength.

For concrete filled hollow sections the full cylinder strength fck may be used owing tothe containment of the concrete.

For concrete filled circular tubes Npl,Rd is modified to take account of the triaxialstresses in the concrete due to its containment, and of the reduction in the allowablestrength of the steel cross-section owing to the induced hoop tension from the concretetriaxial stresses. The modified value of Npl,Rd is subject to two conditions:

(1) λ < 0,5 and(2) e = MEd/NEd ≤ d/10 (where d is the diameter)

The equation for Npl,Rd becomes

Npl,Rd = Aaηafyγa

+ Ac

[1 + ηc

td

fyfck

]fck

γc+ As

fsk

γs(10.59)

where t is the thickness of the tube and ηa and ηc are co-efficients determined asfollows,

ηc = ηc0

(1 − 10

ed

)(10.60)

where ηc0 is given by

ηc0 = 4,9 − 18,5λ + 17(λ)2 ≥ 0 (10.61)

and

ηa = ηa0 + (1 − ηa0)10ed

(10.62)

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hn

hn

Ac As

Ax

(a) Point A

(b) Point B and C

(c) Point D

Neutral axiscentroidal axis

Concrete below axisignored

FIGURE 10.15 Calculation of sectioncapacities

where ηa0 is given by

ηa0 = 0,25(3 + 2λ) ≤ 1,0 (10.63)

For e > d/10, ηc = 0 and ηa = 1,0.

10.4.4.2 Calculation of Mpl,Rd [Point B]

For the typical section given in Fig. 10.15(b), the forces in the flanges and the reinforce-ment cancel out, and thus the tension force in the web must balance the compressionblock in the concrete. It is therefore straightforward to determine the height of theweb hn above the centroidal axis. Equating compressive and tensile forces gives

hn = Npm,Rd

2b fckγc

+ 4tw(

2 fyγa

− fckγc

) (10.64)

where Npm,Rd is given by Eq. (10.67).

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The plastic moment capacity is determined by taking moments about the centroidalaxis of the section. The moment capacity Mpl,Rd is thus given by

Mpl,Rd = Mmax,Rd − Mn,Rd (10.65)

The values of Mmax,Rd and Mn,Rd are given by Eqs (10.71) and (10.68), respectively.

10.4.4.3 Determination of Npm,Rd [Point C]

This can be done by noting that the neutral axis shifts from hn above the centroidal axisto the same distance below it (Fig 10.15(b)). Npm,Rd is determined from horizontalforce equilibrium recognizing that the forces in the steel flanges and reinforcementcancel out, that is, the axial load is carried by the concrete alone.

The area of the concrete Ac is given by

Ac = (b − 2t)(h − 2t) − (4 − π)r2 (10.66)

where r is the corner radius taken equal to t, and Npm,Rd is given by

Npm,Rd = Acfck

γc(10.67)

The moment capacity Mn,Rd is given by

Mn,Rd = Wp,a,nfyγa

+ 0, 5Wp,c,nfck

γc(10.68)

where Wp,a,n and Wp,c,n are the plastic section moduli for the portions of the steel tubeand concrete contained within ± hn, and are given by

Wp,c,n = (b − 2t)h2n (10.69)

and

Wp,a,n = bh2n − Wp,c,n (10.70)

10.4.4.4 Determination of Mmax,Rd [Point D]

For this case when an axial force of 0,5Npm,Rd acts the neutral axis coincides with thecentroidal axis (Fig. 10.15(c)), and thus Mmax,Rd is simply given by the sum of theplastic moment capacities of the reinforcement, the steel section and the concreteabove the centroidal axis. Mmax,Rd is given by

Mmax,Rd = Wpafyγa

+ 0,5Wpcfck

γc(10.71)

where Wpa is the plastic section modulus for the steel section (taken from tables) andWpc is calculated from

Wpc = (b − 2t)(h − 2t)2

4− 2

3r3 − (4 − π)(0,5h − t − r)r2 (10.72)

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Note, the equations have been derived for bending about the major (yy) axis. Forminor axis (zz) bending, h and b are simply interchanged.

10.4.5 Buckling

The non-dimensionalized slenderness ratio λ is defined by

λ =√

Npl,Rk

Ncr(10.73)

where Npl,Rk is the characteristic plastic axial load capacity from Eq. (10.58) orEq. (10.59) with all the materials’ partial safety factors set equal to unity and theeffective buckling load Ncr is given by

Ncr = π2(EI)eff

l2(10.74)

where l is the buckling length with the effective flexural stiffness is given by

(EI)eff = EaIa + KeEcmIc + EsIs (10.75)

where EaIa is the flexural rigidity of the steel section alone, EcmIc is the flexuralrigidity of the concrete, and EsIs is the flexural rigidity of the reinforcement and Ke isa correction factor taken as 0,6. Ecm is given in Table 3.1 of EN 1992-1-1.

Where appropriate, account should be taken of the influence of long-term loading byusing an effective concrete modulus Ec,eff determined from

Ec,eff = Ecm

1 + φtNG,EdNEd

(10.76)

where NEd is the total design normal force and NG,Ed is that portion which is permanentand ϕt is the creep co-efficient determined from EN 1992-1-1.

10.4.6 Design Moments

Second order effects within the column length may be allowed for by increasing thelarger design bending moment determined from a first order analysis by a factor kgiven by

k = β

1 − NEdNcr,eff

≥ 1,0 (10.77)

where the moment ratio factor β for end moments is given by

β = 0,66 + 0,44r ≥ 0,44 (10.78)

and Ncr,eff is an effective buckling load determined using the actual column length andan effective stiffness (EI)eff,II given by

(EI)eff,II = K0(EaIa + Ke,IIEcmIc + EsIs) (10.79)

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where K0 is a calibration factor (=0,9) and Ke,II is a correction factor representing theeffects of cracking in the concrete (=0,5).

Johnson (2004) indicates that where Ncr,eff ≥ 10NEd, then the second order effectsneed considering and there is an additional moment induced by the additionalimperfection.

Thus the design moment MEd is given by

MEd = kendM + kimpNEde0 (10.80)

where kend is magnification factor due to the moment gradient, M is the larger endmoment, e0 is the eccentricity due to imperfections, kimp is the magnification factordetermined from Eq. (10.78) with β = 1,0.

10.4.7 Other Checks and Detailing

10.4.7.1 Local Buckling

For circular hollow sections: d/t ≤ 90ε2;

Rectangular hollow sections: h/t ≤ 52ε;

Partially encased sections: b/tf ≤ 44ε.

10.4.7.2 Cover

For reinforcement, this is governed by the requirements in EN 1992-1-1. For encasedsteel sections, this should be a maximum of 40 mm or b/6, where b is the flange width.

10.4.7.3 Shear

The shear bond between the steel section and the concrete should be checked usingan elastic distribution of forces on the uncracked section with a transmission lengthnot exceeding twice the relevant transverse direction. The values of shear bond shouldnot exceed 0,3 MPa for fully encased sections, 0,55 MPa for circular concrete filledsections, 0,40 MPa for circular concrete filled sections and 0,2 MPa for the flangesonly in partially encased sections. Where necessary shear studs should be used onencased I sections to resist shear.

10.4.7.4 Fire

For concrete filled hollow sections it is essential that two vent holes of 20 mm diametershould be drilled through the steel section at the top and bottom of each storey subjectto a maximum spacing of 5 m (Newman and Simms, 2000). These holes must not bewithin the depth of the floor construction. The purpose of these holes is to allow thebuild up of water vapour to escape whilst the moisture within the concrete is drivenoff in the early stages of heating in the fire.

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It is clear that the method of designing composite columns under uniaxial or biax-ial bending is complex, and leads itself readily to the use of spreadsheets or designcharts. The analysis of given sections to determine their carrying capacity is muchmore straightforward. The first example illustrates the determination of axial carryingcapacity. To avoid duplication of calculations the second and third both use the samesection, one under uniaxial bending about the major axis, and the other with biaxialbending. In each case the loading is considered totally short term, that is, Ec,eff is takenas Ecm. Also any reinforcement is considered negligible and is neglected.

EXAMPLE 10.4 Determination of axial load capacity of a composite column.

Determine the axial load carrying capacity of 4 m effective length 150 × 150 × 8 GradeS355 rolled hollow section filled with Grade C25/30 concrete.

Check h/t:

Actual:

ht

= 1508

= 18,75

Allowable:

52ε = 52 × (235/355)1/2 = 42,3, therefore satisfactory

Determination of Npl,Rd:

Use Eq. (10.58) with the concrete taken at full strength,

Npl,Rd = Aafyγa

+ Ac0,85fck

γc+ As

fsk

γs= 4510

3551,0

+[(150 − 16)2 − (4 − π)82

] 251,5

= 1,899 MN

Determine the load contribution ratio δ from Eq. (10.53):

δ =Aa

fyγa

Npl,Rd=

4510 3551,0 × 10−6

1,899= 0,84

As δ lies between 0,2 and 0,9, the column may be designed as composite.

Use Eq. (10.75) to determine the effective stiffness (EsIs = 0)

Ecm = 22(

fck + 810

)1/3

= 22(

25 + 810

)1/3

= 31,5 GPa

(EI)eff = EaIa + 0,6EcmIc = 210 × 106 × 1510 × 10−8 + 0, 6 × 31, 5 × 106

× (150 − 16)4

12= 3678 kNm2

Determine the Euler critical load Ncr from Eq. (10.74)

Ncr = π2(EI)eff

l2= π2 × 3678

42 = 2269 kN

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Determine Npl,Rk using Eq. (10.58) with the materials’ partial safety factors setequal to 1,

Npl,Rk = Aafyγa

+ Ac0,85fck

γc+ As

fsk

γs= 4510

3551,0

+[(150 − 16)2 − (4 − π)82

] 251,0

= 2,048 MN

Determine the normalized slenderness ratio λ from Eq. (10.73),

λ =√

Npl,Rk

Ncr=√

20482269

= 0,95

This is less than the critical value of 2,0.

The strength reduction factor is determined using buckling curve ‘a’ with α = 0,21,

φ = 0, 5(1 + α(λ − 0, 2) + (λ)2) = 0, 5(1 + 0,21(0,95 − 0,2) + 0,952) = 1,03

χ = 1

φ +√

φ2 − (λ)2= 1

1,03 +√1,032 − 0,952= 0,70

Determine the axial capacity NRd:

NRd = χNpl,Rd = 0,70 × 1899 = 1329 kN

EXAMPLE 10.5 Axial load and uniaxial bending about the major axis.

Determine whether a column having a system length of 4 m fabricated from150 × 100 × 8 Grade S355 RHS filled with Grade C25/30 normal weight concrete cancarry an axial load at ULS of 400 kN and a moment at ULS about the major axis of18 kNm.

Aa = 3710 mm2; Wpl,y = 183 × 103 mm3.

Determine Wpc from Eq. (10.72)

Wpc = (b − 2t)(h − 2t)2

4− 2

3r3 − (4 − π)(0,5h − t − r)r2

= (100 − 2 × 8)(150 − 2 × 8)2

4− 2

383 − (4 − π)

(1502

− 8 − 8)

= 373500 mm3

The area of the concrete Ac is given by Eq. (10.66)

Ac = (b − 2t)(h − 2t) − (4 − π)r2 = (150 − 2 × 8)(100 − 2 × 8) − (4 − π)82

= 11200 mm2

Axial squash capacity Npl,Rd is given by Eq. (10.58):

Npl,Rd = Aafyγa

+ Ac0,85fck

γc+ As

fsk

γs= 3710

3551,0

+ 11200251,5

= 1504 kN

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Determine the contribution ratio, δ from Eq. (10.53):

δ = Asfyd

Npl,Rd=

3710 3551,0 × 10−3

1504= 0,876

This is within the limits for design as a composite column.

Maximum moment capacity Mmax,Rd from Eq. (10.71):

Mmax,Rd = Wpafyγa

+ 0,5Wpcfck

γc= 183000

3551,0

× 10−6 + 0,5 × 373500251,5

× 10−6

= 68,1 kNm

Determine Npm,Rd from Eq. (10.67):

Npm,Rd = Acfck

γc= 11200

251,5

× 10−3 = 187 kN

Npm,Rd

2= 93,5 kN

Determine hn from Eq. (10.64) determine hn:

hn = Npm,Rd

2b fckγc

+ 4tw(

2fyγa

− fckγc

) = 18700

2 × 100 × 251,5 + 4 × 8

(2 355

1,0 − 251,5

) = 7,33 mm

Determine Wp,c,n from Eq. (10.69):

Wp,c,n = (b − 2t)h2n = (100 − 2 × 8)7,332 = 4513 mm3

Determine Wp,a,n from Eq. (10.70):

Wp,a,n = bh2n − Wp,c,n = 100 × 7,332 − 4513 = 860 mm3

Determine Mn,Rd from Eq. (10.68):

Mn,Rd = Wp,a,nfyγa

+ 0,5Wp,c,nfck

γc= 860

3551,0

× 10−6 + 0,5 × 4513251,5

× 10−6

= 0,3 kNm

Determine Mpl,Rd from Eq. (10.65):

Mpl,Rd = Mmax,Rd − Mn,Rd = 68,1 − 0,3 = 67,8 kNm

The values required to plot the interaction diagram are given in Table 10.3.

TABLE 10.3 Values required for major axis interaction diagram for EXAMPLE 10.5.

Point Moment capacity (kNm) Axial capacity (kN)

A (0, N pl,Rd) 0 1504B (Mpl,Rd, 0) 67,8 0C (Mpl,Rd, N pm,Rd) 67,8 187D (Mmax,Rd, 0,5N pm,Rd) 69,1 93,5

Ch10-H5060.tex 12/7/2007 15: 0 page 406


These values are plotted in Fig. 10.15.

Determine the resistance to axial buckling about the major axis:

EaIa = 210 × 106 × 1106 × 10−8 = 2323 kNm2

Ecm = 22(

( fck + 8)10

)1/3

= 31,5 GPa

Ic = (0,150 − 0,016)3(0,100 − 0,016)12

= 16,84 × 10−6 m4

EcdIc = 31,5 × 106 × 16,84 × 10−6 = 530 kNm2

Determine (EI)eff,II from Eq. (10.79):

(EI)eff,II = 0,9(EaIa + 0,5EcmIc) = 0,9(2323 + 0,5 × 530) = 2329 kNm2

Determine Ncr,eff from Eq. (10.74) with (EI)eff replaced by (EI)eff,II:

Ncr,eff = π2(EI)eff,II

l2= π2 × 2329

42 = 1437 kN

NEd = 400 kN > Ncr,eff /10, therefore second order effects need to be considered.

Npl,Rk = (3710 × 355 + 11200 × 25) × 10−3 = 1597 kN.

Determine λ from Eq. (10.73):

λ =√

Npl,Rk

Ncr,eff=√

15971437

= 1,054 > 2,0

Thus the column satisfies the limits for composite design.

Second order effects:

(a) Within the column length:Assuming the column is in single curvature, r = 1,0, so β = 1,0.

kend = β

1 − NEdNcr,eff

= 1,0

1 − 4001437

= 1,386

(b) Due to initial bow:

From Table 6.3 of EN 1994-1-1, for a infilled hollow section, e0 = L/300 = 4/300

As β = 1,0 for initial bow, kimp = 1,386

The design moment MEd is given by Eq. (10.80):

MEd = kendM + kimpNEde0 = 1,386 × 18 + 1,386 × 4004

300= 32,3 kNm

Ch10-H5060.tex 12/7/2007 15: 0 page 407


From the interaction diagram in Fig. 10.15, the moment M400 corresponding to anaxial load of 400 kN is given by

M400 = Mpl,RdNpl,Rd − NEd

Npl,Rd − Npm,Rd= 67,8

1504 − 4001504 − 187

= 56,5 kNm

μd = M400

Mpl,Rd= 56,5

67,4= 0,833

MEd

μdMpl,Rd= MEd

M400= 32,3

56,5= 0,572 < 0, 9

The limiting value of αM is 0,9 as Grade S355 steel is being used. Thus the column istherefore satisfactory.

EXAMPLE 10.6 Axial load and biaxial bending.

Determine whether a column having a system length of 4 m fabricated from150 × 100 × 8 Grade S355 RHS filled with Grade C25/30 normal weight concrete cancarry an axial load at ULS of 350 kN and a moment at ULS about the major axis of 15and 10 kNm about the minor axis.

Two interaction diagrams are required, one for each axis.

For the major axis the interaction diagram is as EXAMPLE 10.5. For the minor axis, theformulae in Eqs (10.64)–(10.73) are used but with b and h interchanged.

Aa = 3710 mm2; Wpl,y = 183 × 103 mm3; Wpl,z = 133 × 103 mm3.

Determine Wpc from Eq. (10.72):

Wpc = (h − 2t)(b − 2t)2

4− 2

3r3 − (4 − π)(0, 5b − t − r) r2

= (150 − 2 × 8)(100 − 2 × 8)2

4− 2

383 − (4 − π)

(1002

− 8 − 8)

= 234200 mm3

From EXAMPLE 10.5, Ac = 11200 mm2, Npl,Rd = 1504 kN and δ = 0,876.

Determine the maximum moment capacity Mmax,Rd from Eq. (10.71):

Mmax,Rd = Wpafyγa

+ 0,5Wpcfck

γc= 133000

3551,0

× 10−6 + 0,5 × 234200251,5

× 10−6

= 49,2 kNm

From EXAMPLE 10.5, Npm,Rd = 187 kN, and 0,5Npm,Rd = 93,5 kN

Determine hn from Eq. (10.64):

hn = Npm,Rd

2h fckγc

+ 4tw(

2fyγa

− fckγc

) = 187000

2 × 150 251,5 + 4 × 8

(2 355

1,0 − 251,5

) = 6,41 mm

Ch10-H5060.tex 12/7/2007 15: 0 page 408


Determine Wpc,n from Eq. (10.69):

Wpc,n = (h − 2t)h2n = (150 − 2 × 8)6,412 = 5506 mm3

Determine Wpa,n from Eq. (10.70):

Wpa,n = hh2n − Wpc,n = 150 × 6,412 − 5506 = 657 mm3

Determine Mn,Rd from Eq. (10.68):

Mn,Rd = Wpa,nfyγa

+ 0,5Wpc,nfck

γc= 657

3551,0

× 10−6 + 0,5 × 5506251,5

× 10−6

= 0,3 kNm

Determine Mpl,Rd from Eq. (10.65):

Mpl,Rd = Mmax,Rd − Mn,Rd = 49,2 − 0,3 = 48,9 kNm

The values required to plot the interaction diagram are given in Table 10.4.

These values are plotted in Fig. 10.16.

TABLE 10.4 Values required for minor axis interaction diagram for EXAMPLE 10.6.

Point Moment capacity (kNm) Axial capacity (kN)

A (0, N pl,Rd) 0 1504B (Mpl,Rd,0) 48,9 0C (Mpl,Rd, N pm,Rd) 48,9 187D (Mmax,Rd, 0,5N pm,Rd) 49,12 93,5

00

200

400

600

800

1000

1200

1400

1600

20 40 60 80

Moment capacity (kNm)

Axi

al c

apac

ity (

kN)

FIGURE 10.16 Major axis M–N interaction diagram (EXAMPLE 10.5)

Ch10-H5060.tex 12/7/2007 15: 0 page 409


Determine the resistance to axial buckling about the minor axis:

EaIa = 210 × 106 × 577 × 10−8 = 1212 kNm2

Ecm = 31,5 GPa (as EXAMPLE 10.5)

Ic = (0,100 − 0,016)3(0,150 − 0,016)12

= 7,11 × 10−6 m4

EcdIc = 31,5 × 106 × 7,11 × 10−6 = 224 kNm2

(EI)eff,II = 0,9(EaIa + 0,5EcmIc) = 0,9(1212 + 0,5 × 224) = 1192 kNm2

Ncr,eff = π2(EI)eff,II

l2= π2 × 1192

42 = 735 kN

NEd = 350 kN > Ncr,eff /10, therefore second order effects need to be considered.

Npl,Rk = Aafy + Acfck = 3710 × 355 × 10−3 + 11200 × 25 × 10−3 = 1597 kN

λ =√

Npl,Rk

Ncr=√

1597735

= 1,474 < 2,0

Second order effects (major yy axis) using the appropriate critical buckling load:


kend,y = β

1 − NEdNcr,eff

= 1,0

1 − 3501437

= 1,322



As β = 1,0 for initial bow, kimp,z = 1,322.

Second order effects (minor zz axis):


kend,z = β

1 − NEdNcr,eff

= 1,0

1 − 350730

= 1,921



As β = 1,0 for initial bow, kimp,z = 1,921.

Ch10-H5060.tex 12/7/2007 15: 0 page 410


00

200

400

600

800

1000

1200

1400

1600

10 20 30 40 50 60Moment capacity (kNm)

Axi

al c

apac

ity (

kN)

FIGURE 10.17 Minor axis M–N interaction diagram (EXAMPLE 10.6)

From the interaction diagram for the major axis in Fig. 10.16, the moment M350

corresponding to an axial load of 350 kN is given by



1504 − 3501504 − 187

= 59,4 kNm

μdy = M350

Mpl,Rd= 59,4

67,4= 0,881

From the interaction diagram in Fig. 10.17, for the minor axis the moment M350

corresponding to an axial load of 350 kN is given by



1504 − 3501504 − 187

= 42,8 kNm

μdy = M350

Mpl,Rd= 42,8

48,9= 0,875

The second order effect due the bow may only be applied on one axis. Use Eq. (10.80)to calculate the design moments.

Two cases, therefore, need considering:

(a) Bow on major axis.

MEd, y = kend, yMSd, y + kimp, yNEde0 = 1,322 × 15 + 1,322 × 3504

300= 26,0 kNm

MEd, z = kend, zMSd, z = 1,921 × 10 = 19,2 kNm

Ch10-H5060.tex 12/7/2007 15: 0 page 411


MEd,y

μdyMpl,y,Rd= 26,0

59,4= 0,438 < 0,9

MEd,z

μdzMpl,z,Rd= 19,2

42,8= 0,45 < 0,9

MEd,y

μdyMpl,y,Rd+ MEd,z

μdzMpl,z,Rd= 26,0

59,4+ 19,2

42,8= 0,887 < 1,0

(b) Bow on minor axis.

MEd,y = kend,yMSd,y = 1,322 × 15 = 19,8 kNm

MEd,z = kend,zMSd,z + kimp,zNEde0 = 1,921 × 10 + 1,921 × 3504

300= 28,2 kNm

MEd,y

μdyMpl,y,Rd= 19,8

59,4= 0,333 < 0,9

MEd,z

μsdzMpl,z,Rd= 28,2

42,8= 0,66 < 0,9

MEd,y

μdyMpl,y,Rd+ MEd,z

μdzMpl,z,Rd= 19,8

59,4+ 28,2

42,8= 0,992 < 1,0

It will be noted that the application of the initial bow to the minor axis is the criticalcase. This is due to both the lower moment capacity and the lower buckling load.

REFERENCES

Bunn, R. and Heywood, M. (2004) Supporting services from structure. Co-Construct (BSRIA).EN 1991-1-1 Eurocode 1: Actions on structures – Part 1–1: General actions – Densities, self-weight,

imposed loads for buildings. CEN/BSI.EN 1991-1-6 Eurocode 1: Actions on structures – Part 1–6: Actions during execution. CEN/BSI.EN 1992-1-1 Eurocode 2: Design of concrete structures – Part 1–1: General rules and rules for buildings.

CEN/BSI.EN 1993-1-1 Eurocode 3: Design of steel structures – Part 1.1: General rules and rules for buildings.

CEN/BSI.EN 1994-1-1 Eurocode 4: Design of composite steel and concrete structures – Part 1.1: General rules and

rules for buildings. CEN/BSI.Evans, H.R. and Wright, H.D. (1988). Steel–concrete composite flooring deck structures, In Steel–

concrete composite structures (ed R. Narayanan). Elsevier.Johnson, R.P. (2004). Composite structures of steel and Concrete (3rd edition). Blackwell Publishing.Johnson, R.P. (2005). Shear connection in beams that support composite slabs – BS 5950 and EN

1994-1-1, Structural Engineer, 83(22), 21–24.Johnson, R.P. and Anderson, D. (2004). Designers’ guide to EN 1994-1-1 Eurocode 4: Design of

composite steel and concrete structures – Part 1.1: General rules and rules for buildings. ThomasTelford.

Johnson, R.P. and May, I.M. (1975). Partial-interaction design of composite beams, The StructuralEngineer, 53(8), 305–311.

Lawson, R.M. and Chung, K.F. (1994). Composite beam design to eurocode 4. Publication 121. SCI.

Ch10-H5060.tex 12/7/2007 15: 0 page 412


Lawson, R.M. and Nethercot, D.A. (1985). Lateral stability of I-beams restrained by profiled sheeting,The Structural Engineer, 63B(1), 1–7, 13.

Lawson, R.M., Mullett, D.L. and Rackham, J.W. (1997). Design of asymmetric Slimflor® beams usingdeep composite decking. Publication 175. SCI.

Martin, L.H. and Purkiss, J.A. (2006). Concrete design to EN 1992. Butterworth-Heinemann.Mottram, J.T. and Johnson, R.P. (1990). Push tests on studs welded through profiled steel sheeting,

The Structural Engineer, 68(10), 187–193.Mullett, D.L. (1992). Slim floor design and construction. Publication 110. SCI.Mullett, D.L. (1997). Design of RHS Slimflor® Edge Beams. Publication 169. SCI.Mullett, D.L. and Lawson, R.M. (1992). Slim floor construction using deep decking. Technical Report

120. SCI.Newman, G.M. and Simms, W.I. (2000). The fire resistance of concrete filled tubes to Eurocode 4.

Technical Report 259. SCI.Oehlers, D.J. and Johnson, R.P. (1987). The strength of shear stud connections in composite beams,

The Structural Engineer, 65B(2), 44–48.Olgaard, J.G., Slutter, R.G. and Fisher, J.W. (1971). Shear strength of stud connectors in lightweight

and normal-weight concrete, Engineering Journal, American Institute of Steel Construction, 8(2),55–64.

Wyatt, T.A. (1989). Design guide on the vibration of floors. Publication 076. SCI.Yam, L.C.P. and Chapman, J.C. (1968). The inelastic behaviour of simply supported composite beams

of steel and concrete, Proceedings of the Institution of Civil Engineers, 41, 651–683.

Ch11-H5060.tex 12/7/2007 14: 22 page 413

C h a p t e r 11 / Cold-formed SteelSections

Thin-walled, cold-formed steel sections are widely used as purlins and rails, the inter-mediate members between the main structural frame and the corrugated roof or wallsheeting in buildings for farming and industrial use (see Fig. 11.1). Trapezoidal sheetingis usually fixed to these members in order to enclose the building. The most commonsections are the zed, channel and sigma shapes, which may be plain or have lips. Thelips are small additional elements provided to a section to improve its efficiency undercompressive loads by enhancing the section ability against local buckling.

Cold-formed steel sections are fabricated by means of folding, press-braking of platesor cold-rolling of coils made from carbon steel. Sheet steel used in cold-formed sectionsis typically 0.9–8 mm thick. It is usually supplied pre-galvanized in accordance withEuropean Standard EN 10142. Galvanizing gives adequate protection for internalmembers or those adjacent to the boundaries of the building envelope. Cold workingof the steel increases its yield strength but also lowers its ductility (see Fig. 11.2). Forexample, a 20% reduction in thickness can increase yield strength by 50% but reduceselongation to as little as 7%, which probably represents the limit of formability forsimple shapes.

The main benefits of using a cold-formed section are not only its high strength-to-weight ratio but also its lightness, which can save costs on transport, erection andthe construction of foundation, and flexibility that the members can be producedin a wide variety of sectional profiles, which can result in more cost effective designs.Examples of the structural use of cold-formed sections include roof and wall members,steel framing, wall partitions, large panels for housing, lintels, floor joists, modularframes for commercial buildings, trusses, space frames, curtain walling, prefabricatedbuildings, frameless steel buildings, storage racking, lighting and transmission towers,motorway crash barriers, etc.

The prime difference between the behaviour of cold-formed sections and hot rolledstructural sections is that cold-formed members involve thin plate elements whichtend to buckle locally under compression. Cold-formed cross-sections are therefore

This chapter is contributed by Long-yuan Li (Aston University) and Xiao-ting Chu (AucklandUniversity of Technology, New Zealand)

Ch11-H5060.tex 12/7/2007 14: 22 page 414

414 • Chapter 11 / Cold-formed Steel Sections

Rai

l sle

eves

Cle

ader

ang

leA

pex

anti-

sag

bars S

peed

fix

anti-

sag

bars

Pur

lin e

xten

sion

s

Pur

lin s

leev

es

Pur

lins

Sid

e an

dga

ble

rails

Dia

gona

l tie

wire

s an

dfix

ing

clea

ts

Inw

ardl

y lip

ped

chan

nel

Cor

ner

deta

il

Sid

e ra

il su

ppor

ts

Rai

l ext

ensi

ons

Eav

es b

eam

FIGURE 11.1 The building using zed and channel sections as purlins and rails(Copy from Albion Sections Ltd design manual by permission)

usually classified as slender because they cannot generally reach their full strengthbased on the amount of material in the cross-section (Rhodes and Lawson, 1992). Thesecondary difference is that cold-formed members have low lateral stiffness and lowtorsional stiffness because of their open, thin, cross-sectional geometry, which givesgreat flexural rigidity about one axis at the expense of low torsional rigidity and lowflexural rigidity about a perpendicular axis. This leads to cold-formed members beingsusceptible to distortional buckling and lateral–torsional buckling.

Ch11-H5060.tex 12/7/2007 14: 22 page 415


Increase of yield strengthdue to strain hardening

Yield point aftercold working

Further loading aftercold working

FractureUltimatestrength

Yield strength

Proportionallimit

Linear regionPerfect plasticity Strain hardening Necking

s

e

FIGURE 11.2 The influence of cold forming on the stress–strain curve of steel

11.1 ANALYTICAL MODEL

Most purlins and rails are laterally restrained by their supported cladding or sheetingeither partially or completely. Hence, it is necessary to consider the influence of thelateral restraints when establishing an analytical model. Also, it is well known that,when a thin-walled beam has one or more cross-sections that are constrained againstwarping, a complex distribution of longitudinal warping stresses can be developed.These warping stresses together with the longitudinal stresses generated by bendingmoments may cause the beam to have local, distortional and lateral–torsional buckling.

Consider a zed section beam that is partially restrained by the sheeting on its upperflange. Without loss of generality, the restraints of the sheeting can be simplifiedby using a translational spring and a rotational spring, both of which are uniformlydistributed along the longitudinal direction of the beam (see Fig. 11.3). Let the originof the coordinate system (x, y, z) be the centroid of the cross-section, with x-axis beingalong the longitudinal direction of the beam, and y- and z-axes taken in the plane of thecross-section, as shown in Fig. 11.3. According to the bending theory of asymmetricbeams (Vlasov, 1961; Oden, 1967) and noticing that for a zed section y- and z-axes usedin Fig. 11.3 are not the principal axes, the constitutive relationships between momentsand generalized strains can be expressed as

My = −EIyd2wdx2 − EIyz

d2v

dx2

Mz = −EIyzd2wdx2 − EIz

d2v

dx2 (11.1)

Ch11-H5060.tex 12/7/2007 14: 22 page 416


Sheeting

Purlin

z,w

Mz

My

ky

kf

Pk Pq

qz

qy

v

w

φ

y, v

(a) (b)

FIGURE 11.3 (a) Purlin-sheeting system and (b) a simplified analysis model

Mω = EIwd2φ

dx2

MT = GITdφ

dx

where My and Mz are the bending moments about y- and z-axes, Mω is the warpingmoment, MT is the twisting moment, E is the modulus of elasticity, G is the shearmodulus, Iy and Iz are the second moments of the cross-sectional area about y- andz-axes, Iyz is the product moment of the cross-sectional area, Iw is the warping constant,IT is the torsion constant, v and w are the y- and z-components of displacement of thecentroid of the cross-section, φ is the angle of twisting.

The above four equations together with three equilibrium equations can be used todetermine seven unknowns (four moments, My, Mz, Mω and MT and three displace-ments, v, w and φ). For the present problem it is convenient to derive the equilibriumequations by using the principle of minimum potential energy. For the partiallyrestrained beam the total potential energy involves the strain energy of the beam,the strain energy of the two springs and the potential of the applied loads, that is

� = Ub + Us + W (11.2)

in which,

Ub = 12

l∫0

[My

(−d2w

dx2

)+ Mz

(−d2v

dx2

)+ Mω

d2φ

dx2 + MTdφ

dx

]dx

= strain energy of the beam

Ch11-H5060.tex 12/7/2007 14: 22 page 417


Us = 12

l∫0

[ky(v + zkφ)2 + kφφ2

]dx = strain energy of the springs

W = −l∫

0

[(v + zqφ)qy + (w − yqφ)qz

]dx = potential of the applied loads

where l is the length of the beam, ky and kφ are the stiffness constants per unit length ofthe translational and rotational springs, qy and qz are the uniformly distributed loadsin y- and z-directions, (−yk, −zk) and (−yq, −zq) are the coordinates of the springand loading points Pk and Pq (see Fig. 11.3), respectively. Substituting Eq. (11.1) intoEq. (11.2) yields

� = 12

l∫0

⎡⎣EIy

(d2wdx2

)2

+ 2EIyzd2v

dx2d2wdx2 + EIz

(d2v

dx2

)2⎤⎦dx

+ 12

l∫0

⎡⎣EIw

(d2φ

dx2

)2

+ GIT

(dφ

dx

)2⎤⎦dx + 1

2

l∫0

[ky(v + zkφ)2 + kφφ2

]dx

−l∫

0

[(v + zqφ)qy + (w − yqφ)qz]dx (11.3)

The following three equilibrium equations can be obtained by the variation of the totalpotential energy with respect to the displacement components, v and w, and the angleof twisting, φ

EIzd4v

dx4 + ky(v + zkφ) + EIyzd4wdx4 = qy

EIyzd4v

dx4 + EIyd4wdx4 = qz

EIwd4φ

dx4 − GITd2φ

dx2 + (z2kky + kφ)φ + zkkyv = qyzq − qzyq (11.4)

For beams that have no restraints, that is, ky = kφ = 0, Eq. (11.4) are simplified to

EIzd4v

dx4 + EIyzd4wdx4 = qy

EIyzd4v

dx4 + EIyd4wdx4 = qz

EIwd4φ

dx4 − GITd2φ

dx2 = qyzq − qzyq (11.5)

Ch11-H5060.tex 12/7/2007 14: 22 page 418


For beams that are fully restrained, that is, v = φ = 0, Eq. (11.4) are simplified to

Rk + EIyzd4wdx4 = qy

EIyd4wdx4 = qz

Rkzk + Mk = qyzk − qzyk (11.6)

where Rk and Mk are the reaction force and reaction moment at the restrained pointPk. Most cold-formed sections are supported by cleats bolted to the web of the sectionas shown in Fig. 11.4. The boundary conditions thus can be assumed as

v = 0 Mz ≈ 0

w = 0 My ≈ 0

φ = 0 Mω ≈ 0 (11.7)

The cleats are designed so that the lower flange of the section does not bear directlyon the rafter, and web crippling problems are avoided. However, the shear or bearingstrength of the connecting bolts is critical to the design.

Governing Eqs (11.4), (11.5) or (11.6) together with boundary conditions (11.7) can beused to determine the displacements and angle of twisting of the beam under the actionof external loads, qy and qz. The bending moments at any place can be calculated usingEq. (11.1). The bending and shear stresses thus can be calculated from the moments

Upslope

Detail ofButted joint

Gap

Cleat

Purlin

FIGURE 11.4 Purlin butted to rafter beam by a cleat

Ch11-H5060.tex 12/7/2007 14: 22 page 419


and shear force as follows:

σx = MzIy − MyIyz

IyIz − I2yz

y + MyIz − MzIyz

IyIz − I2yz

z + E(ω − ω)d2φ

dx2

τmax = 3MT

Lst2 + VAv

(11.8)

where ω is the sectorial coordinate with respect to the shear centre, ω is the averagevalue of ω, Ls is the total length of the middle line section, t is the thickness, V isthe shear force, Av is the shear area. The sectorial coordinates are properties of thecross-section and are calculated as follows (Chu et al., 2004a,b)

ω =s∫

0

hsds and ω = 1Ls

Ls∫0

ωds (11.9)

where hs is the perpendicular distance from a tangent at the point under considerationto the shear centre, and s is the distance from any chosen origin to the same pointmeasured along the middle line of the section.

Equation (11.8) indicates that when warping torsion is involved twisting produces notonly the shear stress but also axial stress. More about warping torsion can be found inthe books of Oden (1967) and Walker (1975). Ye et al. (2004) investigated the influ-ences of restraints on the magnitude and distribution of the axial stress within thecross-section through varying the stiffness constants of two springs. They also usedthe stress pattern obtained from Eq. (11.8) as an input to the finite strip analysis pro-gram and investigated the influence of restraints on the behaviour of local, distortionaland lateral–torsional buckling of channel and zed section beams (Ye et al., 2002).

It is interesting to notice from Eq. (11.6) that, if it is fully restrained the zed sectionbeam bends only in the plane of the web and the bending stress and deflection can becalculated simply based on the bending rigidity of the beam in the plane of the webalthough the section itself is point symmetric, that is,

σx = MyzIy

d4wdx4 = qz

EIy(11.10)

11.2 LOCAL BUCKLING

Cold-formed members are usually very thin, and thus the thin plate elements tend tobuckle locally under compression. The local buckling mode of a cold-formed mem-ber normally involves plate flexure along, with no transverse deformation of a line orlines of intersection of adjoining plates, and can be characterized by a relatively short

Ch11-H5060.tex 12/7/2007 14: 22 page 420


(a) (b)

FIGURE 11.5 Local buckling modes of a zed section (h = 120 mm, b = 75 mm,c = 20 mm, t = 2,5 mm). (a) Under a pure compression (web buckle) and (b) undera pure bending (flange buckle)

half-wavelength of the order of magnitude of individual plate elements, as illustratedin Fig. 11.5.

11.2.1 Elastic Local Buckling Stress

It is known from the stability of plates that, a simply supported rectangular plate maybuckle if it is subjected to compressive loads in the plane of its middle surface. Theelastic critical compressive stress when the plate buckles is expressed as (Bulson, 1970)

σcr,p = kσπ2E12(1 − ν2)

(t

bp

)2

(11.11)

where bp is the width of the plate, ν is Poisson’s ratio and kσ is the buckling coefficientdetermined from

kσ =(

lmbp

)2

+ 2 +(

mbp

l

)2

(11.12)

where l is the length of a plate and m is the number of half waves of the buckling modein the longitudinal direction in which the plate is compressed. Note that, kσ varieswith m. When m = l/bp, kσ has a minimum value of 4, which makes the compressivestress σcr,p critical. This indicates that the buckles approximate to square wave forms,as demonstrated in Fig. 11.5.

Equations (11.11) and (11.12) are only for a plate simply supported on the two longsides and subjected to uniform compressive stresses. For a compression element of

Ch11-H5060.tex 12/7/2007 14: 22 page 421


width bp with different support conditions and/or subjected to non-uniform compres-sive stresses the critical compressive stress can still be calculated using Eq. (11.11)but the buckling coefficient needs to take account the influence of both boundaryconditions and stress pattern. When these factors have been taken into account kσ isexpressed as follows:

For doubly supported compression elements (Table 4.1 in EN 1993-1-5 (2006))

kσ =

⎧⎪⎪⎨⎪⎪⎩

8,21,05 + ψ

0 ≤ ψ ≤ 1

7,81 − 6,29ψ + 9,78ψ2 −1 < ψ < 05,98(1 − ψ)2 −3 < ψ ≤ −1

(11.13a)

For outstand compression elements (support at σ1) (Table 4.2 in EN 1993-1-5 (2006))

kσ =⎧⎨⎩

0,5780,34 + ψ

0 ≤ ψ ≤ 1

1,7 − 5ψ + 17,1ψ2 −1 ≤ ψ < 0(11.13b)

For outstand compression elements (support at σ2) (Table 4.2 in EN 1993-1-5 (2006))

kσ = 0,57 − 0,21ψ + 0,07ψ2 −3 < ψ ≤ 1 (11.13c)

For single edge compression stiffener elements (lips) (cl 5.5.3.2.5 in EN 1993-1-3(2006))

kσ ={

0,5 cp/bp ≤ 0,35

0,5 + 0,83(cp/bp − 0,35)2/3 0,35 < cp/bp ≤ 0,6(11.13d)

where ψ = σ2/σ1 is the ratio of stresses at the two ends of the element (σ1 is thelarger compressive stress, σ2 is the tensile stress or smaller compressive stress, and thecompressive stress is assumed to be positive), cp and bp are the lengths of the middlelines of the lip and flange, respectively.

Equations (11.11) and (11.13) are used to calculate the critical stress of local bucklingof a compression element. For channel and zed sections, the web and lipped flangemay be treated as the doubly supported elements if the lip satisfies the requirementspecified in Section 5.2 in EN 1993-1-3 (2006). Flanges that have no intermediatestiffeners and no edge lips are treated as the outstand elements. When a web or aflange has an intermediate stiffener, the actual width of the element should be takenas the width of the individual part separated by the stiffener. More details for dealingwith elements with intermediate stiffeners can be found in EN 1993-1-3 (2006) andEN 1993-1-5 (2006).

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11.2.2 Post-Buckling Behaviour andthe Calculation of Effective Width

When an element buckles locally it does not necessarily mean that this element willcollapse or loss its ability of carrying loads. In fact, a plate can be allowed to takea considerably increased load beyond initial buckling before any danger of collapseoccurs. This is because the deflections due to buckling are accompanied by stretchingof the middle surface of the plate.

It is not always possible for practical reasons to allow some elements of a structure tobuckle, but if stable buckles can be tolerated, a considerable gain follows in structuralefficiency. For a uniformly compressed rectangular plate, up to the buckling load,the stress distribution is uniform. With increase in load, the central unconstrainedportion of the plate will start to deflect laterally and will therefore not support muchadditional load, whereas the portions close to the supported edges will be constrainedto remain straight and will continue to carry increasing stresses. Figure 11.6 showsthe typical variation of the stress distribution in a plate in pre- and post-bucklingstages. The ultimate strength of the plate is when the maximum stress at the edgesreaches the compressive yield strength of the material. Thus the ultimate load ofthe plate should be calculated based on the stress distribution at failure through thewidth of the plate. The problem, however, is that analysis of the post-buckled plateis a complicated process and no exact closed form results have been obtained forcompressed plates. Therefore, instead of using the stress distribution in the post-buckling range, an alternative approach to assessing the ultimate load of the plate isto use an effective width concept.

fyb fyb

bp

0,5beff 0,5beff

(a) (c)(b)

FIGURE 11.6 Theconcept of effectivewidth. (a) Stressdistribution up tobuckling, (b) stressdistribution at failureand (c) stressdistribution in effectivewidth

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The concept of effective width was originally developed by Von Karman et al. (1932)and calibrated for cold-formed members by Winter (1968). The method assumes thatwhen the ultimate stress is reached, the total load is carried by two fictitious stripsadjacent to the edges of the plate (see Fig. 11.6c), which carry a uniform stress equalto the yield strength of the material, and the central region is unstressed. Obviously,the calculation of the effective width is dependent on the stress distribution at thetime when the plate fails, which is influenced by a number of factors including thepattern of applied compressive stresses and the boundary conditions, relative slen-derness and geometrical imperfections of the plate. Based on large numbers of tests,empirical functions have been developed. In EN 1993-1-5 (2006) the following equa-tions have been recommended for calculating the effective width of a compressionelement.

For doubly supported compression elements (cl 4.4.2 in EN 1993-1-5 (2006))

ρ =

⎧⎪⎪⎨⎪⎪⎩

1,0 λp,red ≤ 0,673

λp,red − 0,055(3 + ψ)

λ2p,red

0,673 < λp,red(11.14a)

For outstand compression elements or single edge compression stiffener elements(cl 4.4.2 in EN 1993-1-5 (2006))

ρ =

⎧⎪⎪⎨⎪⎪⎩

1,0 λp,red ≤ 0,748

λp,red − 0,188

λ2p,red

0,748 < λp,red(11.14b)

in which,

λp,red = λp

√σcom,Ed

fyb/γM0= reduced slenderness

λp =√

fyb

σcr,p= relative slenderness for local buckling

where ρ is the reduction factor to determine the effective width of a compressionelement defined in Tables 11.1 and 11.2, fyb is the basic yield strength of the mate-rial, σcom,Ed (σcom,Ed ≤ fyb/γM0) is the largest compressive stress in the compressionelement, and γM0 is the partial safety factor for resistance of the cross-section.

After the effective widths of individual compression elements have been determined,the effective area, second moments of the effective area and effective section moduluscan be calculated, from which the design values of the resistance to bending momentscan be determined.

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TABLE 11.1 Doubly supported compression elements (ψ = σ2/σ1).

Stress distribution (compression positive) Effective width, beff

ψ = 1beff = ρbbe1 = be2 = 0,5beffs1 s2

be1 bbe2

1 > ψ ≥ 0beff = ρb

be1 = 2beff5 − ψ

be2 = beff − be1

be1 bbe2

s1 s2

0 > ψ

beff = ρbc = ρb1−ψ

be1 = 0,4beffbe2 = 0,6beff

s1

s2

bc b t

be2be1 b

11.3 DISTORTIONAL BUCKLING

Distortional buckling involves both rotation and translation at the corners of the cross-section. Distortional buckling of flexural members such as channel and zed sectionsinvolves rotation of only the compression flange and lip about the flange–web junctionas shown in Fig. 11.7. The web undergoes flexure at the same half-wavelength as theflange buckle, and the compression flange may translate in a direction normal to theweb, also at the same half-wavelength as the flange and web buckling deformations.The elastic distortional buckling stress of cold-formed flexural members can be deter-mined using either analytical methods, such as those suggested in AS/NZS 4600 (1996)and EN 1993-1-3 (2006) or numerical methods, such as the finite strip method (FSM)(Schafer, 1997) and the generalized beam theory (GBT) method (Davies et al., 1993).

11.3.1 The Calculation Method in EN 1993-1-3 (2006)

In EN 1993-1-3 (2006), the design of compression elements with intermediate oredge stiffeners is based on the assumption that the stiffener behaves as a compression

Ch11-H5060.tex 12/7/2007 14: 22 page 425


TABLE 11.2 Outstand compression element (ψ = σ2/σ1).

Stress distribution (compression positive) Effective width, beff

1 > ψ ≥ 0beff = ρb

b beff

s1

s2

0 > ψ

beff = ρbc = ρb1 − ψ

bbeff

s1 bc bt

s2

1 > ψ ≥ 0beff = ρb

beff b

s2

s1

0 > ψ

beff = ρbc = ρb1 − ψ

s1

s2

bc

bbeff

bt

(a) (b)

FIGURE 11.7 Distortional buckling modes of a zed section (h = 120 mm, b = 75 mm,c = 20 mm, t = 2,5 mm) (a) under a pure compression and (b) under a pure bending

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member with continuous partial restraint, with a spring stiffness that depends on theboundary conditions and the flexural stiffness of the adjacent plane elements of thecross-section. The spring stiffness of the stiffener is determined by applying a unitload per unit length to the cross-section at the location of the stiffener, as illustratedin Fig. 11.8, and is determined from

K = 1δ

= Et3

4(1 − ν2)1

b21(b1 + hp)

(11.15)

where δ is the deflection of the centroid of the stiffener due to a unit load, b1 is thehorizontal distance from the web line to the centroid of the effective area of the edgestiffener, and hp is the depth of the web. The elastic critical buckling stress for a longstrut on an elastic foundation of a spring stiffness coefficient K is given by Timoshenkoand Gere (1961) as follows:

σcr,d = π2EIs

Asλ2 + Kλ2

Asπ2 (11.16)

where As and Is are the area and second moment of the effective section of the stiffener,as illustrated in Fig. 11.9 for an edge stiffener, and λ = l/m is the half-wavelength ofthe distortional buckling (l is the member length and m is the number of half waves).For a sufficiently long strut, the critical half-wavelength can be obtained by minimizingthe critical stress as defined by Eq. (11.6) with respect to λ, to give,

λcr = π

(EIs

K

)1/4

(11.17)

bp

be2

b1Cθ

θ

δ

δU

K

U

(a)

(b) (c)

FIGURE 11.8 (a) Distortional buckling model used in EN 1993-1-3, (2006)(b) edge stiffener on an elastic foundation of a spring stiffness coefficient and(c) model used to determine spring stiffness coefficient (copy from EN 1993-1-3(2006) by permission)

Ch11-H5060.tex 12/7/2007 14: 22 page 427


b

K

b

b

aa

be2be1

bP

Cef

f

Cp C

As, ls

b1

FIGURE 11.9 Effective cross-sectionalarea of an edge stiffener in EN1993-1-3 (2006) (copy from EN1993-1-3 (2006) by permission)

Substituting Eq. (11.17) into Eq. (11.16) yields,

σcr,d = 2√

KEIs

As(11.18)

Equation (11.18) is given in EN 1993-1-3 (2006) for calculating the critical stress ofdistortional buckling of the edge stiffener.

The design strength in EN 1993-1-3 (2006) (Section 5.5.3.1) for distortional bucklingis considered by using a reduced thickness of the edge stiffener. The reduction factoris calculated in terms of the relative slenderness as follows,

χd =

⎧⎪⎪⎨⎪⎪⎩

1,0 λd ≤ 0,65

1,47 − 0,723λd 0,65 < λd < 1,38

0,66/λd 1,38 ≤ λd

(11.19)

in which,

λd =√

fyb

σcr,d= relative slenderness for distortional buckling

The procedure for calculating χd can be summarized as follows (Section 5.5.3.2 in EN1993-1-3 (2006)).

• Step 1 Obtain an initial effective cross-section for the stiffener using effectivewidths determined by assuming that the stiffener gives full restraint and thatσcom,Ed = fyb/γM0.

• Step 2 Use the initial effective cross-section of the stiffener to determine the reduc-tion factor for distortional buckling (flexural buckling of a stiffener), allowing forthe effects of the continuous spring restraint.

• Step 3 Optionally iterate to refine the value of the reduction factor for buckling ofthe stiffener; that is, re-calculate the effective widths of the lip and the part of theflange near the lip based on the compressive stress σcom,Ed = χdfyb/γM0 and calculatethe reduction factor again based on the newly calculated effective widths.

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The design value of the resistance to bending moment about the y-axis due to bothlocal and distortional buckling is determined based on the elastic section modulus ofthe effective section,

Mc,Rd = fybWeff,y

γM0(11.20)

where Weff,y is the section modulus of the effective section for bending about y-axis,in which, apart from the effective widths of the web and the part of the flange near toweb are calculated using the local buckling formulae, the effective widths and thick-nesses of the lip and the part of the flange near the lip are calculated using bothlocal and distortional buckling formulae, based on the reduced compressive stress,σcom,Ed = χd fyb/γM0.

11.3.2 The Calculation Methods in AS/NZS 4600 (1996)

In EN 1993-1-3 (2006) the critical stress of distortional buckling of a section is cal-culated based on the model of an edge stiffener on an elastic foundation and theeffect of the distortional buckling on the section properties is taken into account byreducing the thickness of the stiffener. An alternative to determine the critical stressof distortional buckling of a cold-formed steel section is to use AS/NZS 4600 (1996)design code. The elastic distortional buckling formulae for channel and zed sectionsin AS/NZS 4600 (1996) are based on a simple flange buckling model where the flangeis treated as a thin-walled compression member, as shown in Fig. 11.10, undergoingflexural–torsional buckling (Lau and Hancock, 1987; Hancock, 1997). The rotationalspring stiffness kθ represents the flexural restraint provided by the web which is in flex-ure, and the translational spring stiffness kx represents the resistance to translationalmovement of the section in the buckling mode. The model includes a reduction in theflexural restraint provided by the web as a result of the compressive stress in the web.

Lau and Hancock (1987) showed that the translational spring stiffness has no signifi-cant influence on results and thus is assumed to be zero. The rotational spring stiffnessand the critical stress at distortional buckling are given as

kθ = 2Et3

5,46(hp + 0,06λ)

[1 − 1,11σcr,d

Et2

(h4

pλ2

12,56λ4 + 2,192h4p + 13,39λ2h2

p

)](11.21)

Kx KθShear centre

x�

y�

x

y

FIGURE 11.10 Flangeelastically restrained alongflange-web junction inAS/NZS 4600 (1996)

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σcr,d = E2 Af

[(α1 + α2) −

√(α1 + α2)2 − 4α3

](11.22)

in which,

α1 = η

β1

(Ixfb2

p + 0,039Jfλ2)

+ kθ

β1ηE

α2 = η

(Iyf + 2

β1ybpIxyf

)

α3 = η

(α1Iyf − η

β1I2xyfb

2p

)

β1 = x2 +(

Ixf + Iyf

Af

)

λ = 4,80

(Ixfb2

php

2t3

)1/4

η =(π

λ

)2

where Af is the full cross-sectional area of the compression flange and lip, Ixf and Iyf

are the second moments of the area Af about x′- and y′-axes, respectively, where thex′- and y′-axes are located at the centroid of area Af with x′-axis parallel with flange,Ixyf is the product moment of the area Af about x′- and y′-axes, Jf is the St Venanttorsion constant of the area Af , x and y are the distances from the flange-web junctionto the centroid of area Af in the x′- and y′-directions, respectively. Due to the couplingof σcr,d and kθ in Eqs (11.21) and (11.22), Hancock (1997) suggested that kθ can becalculated based on an initial σcr,d obtained by assuming kθ = 0 and after then σcr,d canbe calculated based on the obtained kθ value. In the iteration, if kθ < 0, kθ should becalculated using σcr,d = 0.

The elastic critical moment for distortional buckling is calculated based on the criticalbuckling stress as follows,

Mcr,d = σcr,dWy (11.23)

where Wy is the elastic section modulus of the gross cross-section for the extremecompression fibre. The design value of the resistance to bending moment about y-axisdue to distortional buckling which involves rotation of the compression flange and lipabout the flange-web junction is calculated as follows:

Mc,Rd =

⎧⎪⎨⎪⎩

Mc 0 ≤ kθ

McWeff,y

Wykθ < 0

(11.24)

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in which,

Mc =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Myield

(1 − Myield

4Mcr,d

)Weff,y

Wy0,5Myield < Mcr,d

Myield

⎡⎣0,055

(√Myield

Mcr,d− 3,6

)2

+ 0,237

⎤⎦Weff,y

WyMcr,d ≤ 0,5Myield

where Myield = fybWy is the moment causing initial yield at the extreme compressionfibre of the gross section.

A comparison of the critical stresses of distortional buckling by using EN 1993-1-3(draft version of 2001) and AS/NZS 4600 (1996) with experimental data wasmade by Kesti and Davies (1999). It was found that Lau and Hancock’s ana-lytical expressions give a good prediction of the distortional buckling stress. Themethod given in EN 1993-1-3 does not correlate as well as Lau and Hancock’smethod. The error in the distortional buckling stress could lead to a consequen-tial error in the effective cross-sectional area depending on the distortional bucklingstress level.

11.4 LATERAL–TORSIONAL BUCKLING

In practice, purlins and rails are usually used together with their supported cladding orsheeting, and thus they are generally considered to be restrained against lateral deflec-tions perpendicular to the line of action of the loading. If a beam is fully restrained onits translational and rotational degrees neither will the beam rotate nor deflect later-ally. However, if the cladding or sheeting is not strong enough then it is possible forthe beam to become unstable and for very large lateral deflections to occur at a criticalvalue of the applied load. This type of behaviour is called lateral–torsional buckling.Chapter 5 has discussed the lateral–torsional buckling of unrestrained beams. In thissection it is to deal with the lateral–torsional buckling of zed section beams with partialrestraints from the sheeting. For the lateral–torsional buckling of partially restrainedchannel section beams readers can see the work of Chu et al. (2004).

11.4.1 Critical Moment of Lateral–Torsional Buckling

The model presented here for analysing the lateral–torsional buckling of partiallyrestrained beams was originally developed by Li (2004) and lately expanded by Chuet al. (2004b), which is similar to that described in Section 11.1. Assume that the dis-placements and moments in a state of equilibrium are (v,w,φ) and (My,Mz,Mω,MT).Now let vb and wb be the y- and z-components of the buckling displacement of thecentroid of the cross-section and φb be the buckling angle of twisting of the sec-tion. Assuming that the displacements in pre-buckling are very small, the increase

Ch11-H5060.tex 12/7/2007 14: 22 page 431


of the strain energy of the system due to the lateral–torsional buckling thus can beexpressed as

U2 = 12

l∫0

⎡⎣EIy

(d2wb

dx2

)2

+ 2EIyzd2vb

dx2d2wb

dx2 + EIz

(d2vb

dx2

)2⎤⎦dx

+ 12

l∫0

⎡⎣EIw

(d2φb

dx2

)2

+ GIT

(dφb

dx

)2⎤⎦dx

+12

l∫0

[ky(vb + zkφb)2 + kφφ2b]dx (11.25)

On the other hand, the lateral–torsional buckling leads to a decrease of the potentialof the pre-buckling moments and loads, which can be expressed as (Chu, 2004 Li,2004; Chu et al., 2004b)

W2 =l∫

0

[(Mzφb)

d2wb

dx2 − (Myφb) d2vb

dx2 + 12

Mω

(dφb

dx

)2]

dx

+l∫

0

12

[(qyyq + qzzq)φ2b]dx (11.26)

If the net change of the total potential is positive for any of possible buckling displace-ments, that is, U2 > W2, then the equilibrium of the pre-buckling state is said to bestable because the generation of the buckling displacements requires an energy inputinto the system. On the other hand, if the net change of the total potential is negative,that is, U2 < W2, then the equilibrium of the pre-buckling state is said to be unstablebecause the buckling displacements can be generated without any input of energy. Acritical state between stable and unstable equilibria from which the critical load of thelateral–torsional buckling can be determined is

U2 = W2 (11.27)

Equation (11.27) is an eigenvalue type equation. For given reference loads the smallesteigenvalue and corresponding eigenvector calculated from Eq. (11.27) represent thecritical loading factor and corresponding buckling mode.

The above model can be applied directly to the purlins with intermediate lateralrestraints such as provided by anti-sag bars if the pre-buckling moments are calculatedbased on the same model and the buckling displacements satisfy the displacementrestraint conditions at the places where the anti-sag bars are placed. It is difficult toachieve closed form solutions of Eq. (11.27). In most cases only numerical solutionscan be obtained. A general numerical computation procedure has been described by Li(2004) and Chu et al. (2004b) to obtain the critical buckling load, in which the following

Ch11-H5060.tex 12/7/2007 14: 22 page 432


piecewise cubic spline functions are used to construct the displacement fields beforeand during buckling

⎧⎪⎨⎪⎩

v(x)w(x)φ(x)

⎫⎪⎬⎪⎭ =

∑⎡⎢⎣Ni(x) 0 0

0 Ni(x) 00 0 Ni(x)

⎤⎥⎦⎧⎪⎨⎪⎩

δvi

δwi

δφi

⎫⎪⎬⎪⎭ (11.28)

⎧⎪⎨⎪⎩

vb(x)wb(x)φb(x)

⎫⎪⎬⎪⎭ =

∑⎡⎢⎣Ni(x) 0 0

0 Ni(x) 00 0 Ni(x)

⎤⎥⎦⎧⎪⎪⎨⎪⎪⎩

δbvi

δbwi

δbφi

⎫⎪⎪⎬⎪⎪⎭ (11.29)

where Ni(x) is the spline interpolation function at node i and {δvi, δwi, δφi} and{δb

vi, δbwi, δ

bφi} are the nodal displacement vectors before and during buckling. By using

the displacement expressions (11.28) and (11.29), the equilibrium Eq. (11.4) andbuckling Eq. (11.27) can be simplified into the following algebraic matrix equations

⎡⎢⎣Kvv Kvw Kvφ

Kvw Kww 0Kvφ 0 Kφφ

⎤⎥⎦⎧⎪⎨⎪⎩

δv

δw

δφ

⎫⎪⎬⎪⎭ =

⎧⎪⎨⎪⎩

Fv

Fw

Fφ

⎫⎪⎬⎪⎭ (11.30)

⎡⎢⎣Kvv Kvw Kvφ

Kvw Kww 0Kvφ 0 Kφφ

⎤⎥⎦⎧⎪⎨⎪⎩

δbv

δbw

δbφ

⎫⎪⎬⎪⎭ = qcr

qref

⎡⎢⎣

0 0 K0vφ

0 0 K0wφ

K0vφ K0

wφ K0φφ

⎤⎥⎦⎧⎪⎨⎪⎩

δbv

δbw

δbφ

⎫⎪⎬⎪⎭ (11.31)

where Kij (i, j = v,w,φ) is the stiffness matrix, K0ij (i, j = v,w,φ) is the geometric stiffness

matrix, δj ( j = v,w,φ) is the nodal displacement vector, Fj ( j = v,w,φ) is the nodal forcevector, δb

j ( j = v,w,φ) is the nodal vector of buckling displacements, (qcr/qref ) representsthe scale factor between the critical and referenced loads.

As an example, Fig. 11.11 shows the critical load factors of simply supported zedpurlin beams (web depth h = 202 mm, flange width b = 75 mm, lip length c = 20 mmand thickness t = 2,3 mm), with zero, one and two anti-sag bars, subjected to a uni-formly distributed uplift load. It can be seen from the figure that the lateral restrainthas remarkable influence on the lateral–torsional buckling of the beam with no anti-sag bars. The influence is found to decrease with the increase of the beam length.Interestingly, when the beam has one or two anti-sag bars, the influence of the lateralrestraint on the lateral–torsional buckling becomes almost negligible. This impliesthat the anti-sag bar not only has the ability of increasing the critical load but also canreduce the influence of the lateral restraint provided by cladding. Practically, purlinsare often used as continuous beams over two or more spans, in which case the bound-ary conditions of the beam can be regarded as simply supported at one end and fixedat the other end. Fig. 11.12 shows the critical load factors of the zed purlin beam withthis kind of boundary conditions. The results show that there is a significant increasein critical load when the purlin has a fixed boundary condition.

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3000 4000 5000 6000 7000 8000 9000 10 000

Beam length, mm

5

4.5

4

3.5

3

2.5

2

1.5

1

0

0.5

Crit

ical

load

ing

fact

orky�0, 2 bars

ky�0, 1 bar

ky��, 2 bars

ky��, no bars

ky��, 1 bar

ky�0, no bars

FIGURE 11.11 Lateral–torsional buckling of simply supported zed section beams(h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) subjected to uniformlydistributed uplift loading, with various different restraints

3000 4000 5000 6000 7000 8000 9000 10 000

Beam length, mm

5

4.5

4

3.5

3

2.5

2

1.5

1

0

0.5

Crit

ical

load

ing

fact

or ky�0, 2 bars

ky�0, 1 bar

ky��, 2 bars

ky��, no bars

ky��, 1 bar

ky�0, no bars

FIGURE 11.12 Lateral–torsional buckling of pinned-fixed zed section beams(h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm) subjected to uniformlydistributed uplift loading, with various different restraints

Ch11-H5060.tex 12/7/2007 14: 22 page 434


11.4.2 Buckling Resistance Moment ofBeams Subject to Bending

The critical load calculated from the lateral–torsional buckling is based on an idealizedmodel in which the beam has no geometrical imperfections, the cross-section of thebeam does not deform, and the beam does not buckle either locally or distortionallybefore the lateral–torsional buckling occurs. When designing a real member, however,these factors should be taken into account. The method for determining the designbuckling resistance moment for the lateral–torsional buckling of cold-formed sectionbeams is the same as that used for other steel section members (Section 6.3.2.2 in EN1993-1-1 (2005)), that is,

Mb,Rd = χLTWeff,yfyb

γM1(11.32)

in which,

χLT = 1

φLT +(φ2

LT − (λLT)2)1/2 = reduction factor for lateral–torsional buckling (χLT ≤ 1,0)

φLT = 0, 5[1 + αLT(λLT − 0,2

) + (λLT)2] = factor used to calculate the reduction

factor

λLT =(

Mc,Rd

Mcr,LT

)1/2

= relative slenderness for lateral–torsional buckling

where αLT = 0,34 is the imperfection factor, γM1 = 1,0 is the partial factor for resistanceof members to instability and Mcr,LT is the elastic critical moment of the gross cross-section for lateral–torsional buckling about the main axis.

11.5 CALCULATION OF DEFLECTIONS

The deflections of channel and zed section beams under uniformly distributed trans-verse loads can be evaluated using the analytical model presented in Section 11.1,provided that the load does not exceed the critical loads of local and distortionalbuckling. For the evaluation of deflections at loads greater than any critical loadthe influence of the local buckling and/or distortional buckling must be taken intoaccount. The precise analysis of the post-buckling behaviour of a cold-formed sectionbeam, however, is very difficult. A simple approach is to assume that the relationshipsbetween the load and deflection are linear for both pre- and post-buckling analyses asillustrated in Fig. 11.13. In pre-buckling region, the deflections can be calculated usingsimple beam theory and the gross section properties of the beam, since the beam isfully effective before buckling. In post-buckling region, the deflections can be evalu-ated using simple beam theory and the reduced section properties of the beam, sincethe beam is not fully effective after buckling. This assumption is only approximatesince, in reality, the line in post-buckling region is not a straight line, but the errorsintroduced in the approximation are acceptable and conservative when fully reduced

Ch11-H5060.tex 12/7/2007 14: 22 page 435


MRd

�Rd

Mcr

�cr

M

D

FIGURE 11.13 Simplified model used for the calculation of deflections

section properties are used. Note that, in Fig. 11.13 � in the pre-buckling region and(� − �cr) in the post-buckling region are linearly proportional to M and (M − Mcr),respectively. Therefore, the deflection of the beam, �, at an applied moment, M , canbe expressed as

� =

⎧⎪⎪⎨⎪⎪⎩

�crM

McrM ≤ Mcr

�cr

(1 + M − Mcr

Mcr

IIeff

)Mcr < M ≤ MRd

(11.33)

where Mcr and MRd are the critical moment and moment resistance of the beam,�cr is the deflection of beam when it buckles, I is the second moment of the grosscross-sectional area, and Ieff is the second moment of the effective cross-sectional area.

11.6 FINITE STRIP METHODS

The FSM was originally developed by Cheung (1976) and it can be considered asa specialization of the finite element method (Zienkiewicz and Taylor, 2000). Themethod is mainly applied to structures whose geometries do not vary with at least oneof the coordinate axes. The approach of the FSM is considered particularly favourablewhen dealing with the initial buckling or natural frequency characteristics of thin-walled prismatic structures. In the FSM, the prismatic structure is discretized into anumber of longitudinal strips and the displacement fields associated with each strip

Ch11-H5060.tex 12/7/2007 14: 22 page 436


(a) (b)

FIGURE 11.14 (a) The finite element analysis and (b) the finite strip analysis

vary sinusoidally along the strip length and algebraically across the strip width. Similarto the finite element method, shape functions are also used to define the variationof displacement fields along the strip width but they are only the functions of thecross-section coordinates of the strip (see Fig. 11.14).

The use of the FSM for understanding and predicting the behaviour of cold-formedsteel members was pioneered by Lau and Hancock in Australia (1986, 1989). Theymodified the stiffness matrices derived in Cheung’s book and created a commercialcomputer program for solution of the elastic buckling problem of open thin-walledmembers via finite strip called Thin-Wall. Similar programs were developed by Lough-lan (1993) in the UK and by Schafer (1997) in USA. The program developed bySchafer is available in internet (http://www.ce.jhu.edu/bschafer/cufsm) and is particu-larly friendly to use. The codes were written in Matlab language and thus can easilybe modified by users.

11.6.1 Element Stiffness Matrix of the Strip

Consider a strip shown in Fig. 11.15, in which the local coordinate system is definedas that, the x- and y-axes are the two axes within the plane of the strip and the z-axisis normal to the plane of the strip. The three components of buckling displacementof the strip at a point (x, y) can be expressed in terms of the nodal displacements asfollows

{v(x, y)u(x, y)

}=⎡⎢⎣ sin

mπxa

0

0 cosmπx

a

⎤⎥⎦⎡⎢⎣1 − y

b0

yb

0

0 1 − yb

0yb

⎤⎥⎦⎧⎪⎪⎪⎨⎪⎪⎪⎩

v1m

u1m

v2m

u2m

⎫⎪⎪⎪⎬⎪⎪⎪⎭

(11.34)

Ch11-H5060.tex 12/7/2007 14: 22 page 437


w(x, y) = sinmπx

a

[1 − 3y2

b2 + 2y3

b3 , y − 2y2

b+ y3

b2 ,3y2

b2 − 2y3

b3 ,y3

b2 − y2

b

]⎧⎪⎪⎪⎨⎪⎪⎪⎩

w1m

θ1m

w2m

θ2m

⎫⎪⎪⎪⎬⎪⎪⎪⎭

(11.35)

where u(x,y) and v(x,y) are the plane displacements, w(x,y) is the deflection, (u1m, v1m,w1m, θ1m) and (u2m, v2m, w2m, θ2m) are the nodal displacements associated with wavenumber m, a and b are the length and width of the strip, respectively. The assumeddisplacement functions satisfy only the simply supported conditions at the two endsides of the strip. The strain energy of the strip is given as follows

U2 = Et2(1 − ν2)

a∫o

b∫o

[(εx + εy

)2 − 2(1 − ν)

(εxεy − γ2

xy

4

)]dx dy

+ Et3

24(1 − ν2)

a∫o

b∫o

[(κx + κy

)2 − 2(1 − ν)(κxκy − κ2

xy

)]dx dy (11.36)

The membrane and bending strains in Eq. (11.36) are defined as follows,

εx = ∂u∂x

εy = ∂v

∂yγxy = ∂u

∂y+ ∂v

∂x

κx = −∂2w∂x2 κy = −∂2w

∂y2 κxy = − ∂2w∂x ∂y

(11.37)

The element stiffness matrix can be obtained by substituting Eqs (11.34) and (11.35)into Eq. (11.37) and then into Eq. (11.36), that is,

U2 = 12{δ}T

m[K]m{δ}m (11.38)

in which,

{δ}m = u1m, v1m, u2m, v2m, w1m, θ1m, w2m, θ2mT = element nodal vector

[K]m =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Kuv11 Kuv

12 Kuv13 Kuv

14Kuv

22 −Kuv14 Kuv

24Kuv

11 −Kuv12

Kuv22

[0]4x4

symmetric

Kwθ11 Kwθ

12 Kwθ13 Kwθ

14Kwθ

22 −Kwθ14 Kwθ

24Kwθ

11 −Kwθ12

Kwθ22

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= element stiffness matrix

Kuv11 = 12D

t2

[12

ab

+ 1 − ν

12ba

(mπ)2]

Kuv12 = 12D

t2

[3ν − 1

8(mπ)

]

Ch11-H5060.tex 12/7/2007 14: 22 page 438


Kuv13 = 12D

t2

[−1

2ab

+ 1 − ν

24ba

(mπ)2]

Kuv14 = 12D

t2

[1 + ν

8(mπ)

]

Kuv22 = 12D

t2

[1 − ν

4ab

+ 16

ba

(mπ)2]

Kuv24 = 12D

t2

[−1 − ν

4ab

+ 112

ba

(mπ)2]

Kwθ11 = D

a2

[1370

ba

(mπ)4 + 65

ab

(mπ)2 + 6(a

b

)3]

Kwθ12 = D

a

[11420

(ba

)2

(mπ)4 + 1 + 5ν

10(mπ)2 + 3

(ab

)2]

Kwθ13 = D

a2

[9

140ba

(mπ)4 − 65

ab

(mπ)2 − 6(a

b

)3]

Kwθ14 = D

a

[− 13

840

(ba

)2

(mπ)4 + 110

(mπ)2 + 3(a

b

)2]

Kwθ22 = D

[1

210

(ba

)3

(mπ)4 + 215

ba

(mπ)2 + 2ab

]

Kwθ24 = D

[− 3

840

(ba

)3

(mπ)4 − 130

ba

(mπ)2 + ab

]

D = Et3

12(1 − ν2)

11.6.2 Element Geometric Stiffness Matrix of the Strip

The geometric stiffness matrix of a strip subjected to linearly varying edge tractioncan be derived by considering the change of the potential of the in-plane forces duringbuckling. Similar to the approach described in Section 11.4, the change of the potentialof the in-plane forces can be expressed as,

W2 = t2

a∫o

b∫o

[σ1 − (σ1 − σ2)

yb

][(∂u∂x

)2

+(

∂v

∂x

)2

+(

∂w∂x

)2]

dx dy (11.39)

where σ1 and σ2 are the compressive stresses at nodes 1 and 2 (see Fig. 11.15).Substituting Eqs (11.34) and (11.35) into Eq. (11.39) yields

W2 = 12{δ}T

m[Kg]

m{δ}m (11.40)

Ch11-H5060.tex 12/7/2007 14: 22 page 439


a

s2s1

b

a

v2

w1 w2

θ1

θ2

x

y

zu1

v1

u2

FIGURE 11.15 Local coordinates, degrees of freedom and stress distribution ina strip

in which,

[Kg]m = b(mπ)2

1680a

[[Kuv

g ] [0]4x4

[0]4x4 [Kwθg ]

]= element geometric stiffness matrix

[Kuvg ] =

⎡⎢⎢⎢⎣

70(3T1 + T2) 0 70(T1 + T2) 070(3T1 + T2) 0 70(T1 + T2)

70(T1 + 3T2) 0symmetric 70(T1 + 3T2)

⎤⎥⎥⎥⎦

[Kwθg ] =

⎡⎢⎢⎢⎣

8(30T1 + 9T2) 2b(15T1 + 7T2) 54(T1 + T2) −2b(7T1 + 6T2)b2(5T1 + 3T2) 2b(6T1 + 7T2) −3b2(T1 + T2)

24(3T1 + 10T2) −2b(7T1 + 15T2)symmetric b2(3T1 + 5T2)

⎤⎥⎥⎥⎦

For a member composed of multiple strips the global stiffness matrix and global geo-metric stiffness matrix can be obtained by the assembly of element stiffness matricesand element geometric stiffness matrices, that is

[Km] =∑

k

[K]m [Kgm] =∑

k

[Kg]m (11.41)

where the index k denotes the k-th element. The summation implies proper coordinatetransformations and correct addition of the stiffness terms in the global coordinatesand degrees of freedom. The elastic buckling problem is a standard eigenvalue problemof the following form:

[Km]{δm} = λm[Kgm]{δm} (11.42)

where eigenvalues, λm, and eigenvectors, {δm}, are the buckling loads and correspond-ing buckling modes.

Ch11-H5060.tex 12/7/2007 14: 22 page 440


11.6.3 Finite Strip Solution Methods

The finite strip analysis employs single wave functions and thus can only be applied tothe case where the longitudinal stresses do not vary along the longitudinal axis. Notethat both [Km] and [Kgm] are the functions of the strip length, a, and wave number, m.The buckling loads and buckling modes solved from Eq. (11.42) are also the functionsof them. Since only the critical buckling load, which is the smallest eigenvalue forany of possible wave numbers, is of interest, the strip length and wave number arenot independent each other. There are two quick ways to find the critical bucklingload. One is to let a equal to the real length of the strip and solve the problem forseveral numbers of m. The other is to let m = 1 and solve the problem for a number oflengths. The former provides the results only for the given length of the strip, whereasthe latter provides a complete picture of the critical buckling loads and modes atvarious different half-wavelengths, which has clear physical meanings. For this reasonthe latter method is often used.

Figure 11.16 shows the buckling curves of a laterally restrained (both displacementand rotation) zed section beam (h = 202 mm, b = 75 mm, c = 20 mm and t = 2,3 mm)subjected to pure bending, in which the three local minima represent the local, dis-tortional and secondary distortional buckling. The secondary distortional buckling issometimes called lateral–distortional buckling to distinguish it from flange distortionalbuckling. The secondary distortional buckling exists only when the section is restrained

3

2.5

2

1.5

1

0

0.5

Cri

tica

l lo

adin

g f

acto

r

Critical load curveBuckling curve m = 1Buckling curve m = 2–8

102 103

565 mm

110 mm

3000mm

Beam length, mm

FIGURE 11.16 Buckling curves of a simply supported zed section beam restrained atthe junction between web and tension flange subjected to pure bending (h = 202 mm,b = 75 mm, c = 20 mm, t = 2,3 mm) (copy from Chu et al. (2006) by permission)

Ch11-H5060.tex 12/7/2007 14: 22 page 441


laterally. Note that the buckling curves only provide information about how and whenthe beam buckles. The actual critical load of the beam for a given length should betaken as the lowest value from all of the buckling curves of the same length, and thiscritical load curve is the solid line plotted in Fig. 11.16.

The original FSM can only deal with the buckling problem of members subjected toeither pure compression or pure bending. Recently, Chu et al. (2005, 2006) modifiedthe geometric stiffness matrix in the FSM by allowing for the variation of the pre-buckling stress along the longitudinal axis, thus leading to a semi-analytical FSM whichis able to deal with the buckling problem of cold-formed section beams subjected touniformly distributed transverse loading. As an example, Fig. 11.17 shows the criticalload curve of the restrained zed section beam (h = 202 mm, b = 75 mm, c = 20 mm andt = 2,3 mm) under uniformly distributed loading. In order to compare the difference incritical load between pure bending and uniformly distributed loading, the results of thebeam under pure bending is also superimposed in the figure. It is evident that the crit-ical load associated with uniformly distributed loading is significantly higher than thatarising from pure bending although, for both local and distortional buckling, the differ-ences between the two loading cases decrease with the beam length. For example, for a5 m long beam, the critical load of a uniformly distributed loading beam is 12% higherthan that of the pure bending beam compared to 20% higher for a 2 m long beam.

4

3.5

3

2.5

2

1.5

1

0

0.5

Cri

tica

l lo

adin

g f

acto

r

110 mm

565 mm

3000mm

Uniformly distributed load

Pure bending

102 103

Beam length, mm

FIGURE 11.17 Critical load curves of a simply supported zed section beamrestrained at the junction between web and tension flange subjected to a uniformlydistributed uplift load and a pure bending (h = 202 mm, b = 75 mm, c = 20 mm,t = 2,3 mm) (copy from Chu et al. (2006) by permission)

Ch11-H5060.tex 12/7/2007 14: 22 page 442


11.7 DESIGN METHODS FOR BEAMS PARTIALLY RESTRAINED BYSHEETING

Purlins and rails are usually used together with their supported trapetzoidal sheeting.Thus it is generally assumed that the sheeting takes the load in the plane of the sheetingand the purlin takes the load normal to the plane of the sheeting, that is, in the plane ofweb. Also, the purlin may be regarded as being laterally restrained in the plane of thesheeting and partially restrained in twisting if the trapetzoidal sheeting is connected toa purlin and the connection meets the following condition (cl 10.1.1.6 in EN 1993-1-3(2006))

S ≥ 70h2

(π2EIw

l2+ GIT + π2h2EIz

4l2

)(11.43)

where S is the portion of the shear stiffness provided by the sheeting for the examinedmember connected to the sheeting at each rib (if the sheeting is connected to a purlinevery second rib only, then S should be substituted by 0,2S), h is the web depth and lis the span length. The partial torsional restraint may be represented by a rotationalspring with a spring stiffness CD, which can be calculated based on the stiffness of thesheeting and the connection between the sheeting and the purlin, as follows,

1CD

= 1CD,A

+ 1CD,C

(11.44)

where CD,A is the rotational stiffness of the connection between the sheeting and thepurlin and CD,C is the rotational stiffness corresponding to the flexural stiffness ofthe sheeting. Both CD,A and CD,C are specified in Section 10.1.5.2 in EN 1993-1-3(2006).

The restraints of the sheeting to the purlin have important influence on the bucklingbehaviour of the purlin. Fig. 11.18 shows the buckling curves of a simply supportedzed purlin beam (h = 202 mm, b = 75 mm, c = 20 mm and t = 2,3 mm) with various dif-ferent lateral restraints applied at the junction between the web and the compressionflange when subjected to a pure bending. The figure shows that, when the transla-tional displacement of the compression flange is restrained the purlin does not bucklelateral torsionally. On the other hand, when the rotation of the compression flange isrestrained the critical stresses of local buckling and distortional buckling are increasedquite significantly. However, when the restraints are applied at the junction betweenthe web and the tension flange, it is only the rotational restraint that influences thelateral–torsional buckling of the purlin (see Fig. 11.19).

Similar to the buckling behaviour, the lateral restraints also have considerable influ-ence on the bending behaviour of the purlin. It has been found from both finiteelement analyses and experiments that, the bending behaviours are different whenthe restraints are applied at tension and compression flanges, and the free flange andthe restrained flange are bent differently, particular when the free flange is under

Ch11-H5060.tex 12/7/2007 14: 22 page 443


102 1030

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Half-wavelength, mm

Rat

io o

f cr

itica

l buc

klin

g st

ress

to

yiel

d st

reng

th

No restraintTranslational restraintRotational restraintFull restraints

FIGURE 11.18 Buckling curves of a simply supported zed section beam withdifferent restraint applied at the junction between web and compression flangesubjected to pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm)

102 1030

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Half-wavelength, mm

Rat

io o

f cr

itica

l buc

klin

g st

ress

to

yiel

d st

reng

th

No restraintTranslational restraintRotational restraintFull restraints

FIGURE 11.19 Buckling curves of a simply supported zed section beam withdifferent restraint applied at the junction between web and tension flange subjectedto pure bending (h = 202 mm, b = 75 mm, c = 20 mm, t = 2,3 mm)

compression (Lucas et al., 1997a, b; Vrany, 2002). This led to development of differ-ent treatments for sections when subjected to gravity loading and uplift loading, since,for instance, for a simply supported beam, the free flange is in tension for gravityloading but in compression for uplift loading. In EN 1993-1-3 (2006), the stress in the

Ch11-H5060.tex 12/7/2007 14: 22 page 444


restrained flange is calculated based on the bending moment in the plane of web asfollows (assuming there is no axial force),

σmax,Ed = My,Ed

Weff,y≤ fy

γM0(11.45)

where fy is the yield strength. While the stress in the free flange is calculated basedon not only the bending moment in the plane of web but also the bending moment inthe free flange due to the equivalent lateral load acting on the free flange caused bytorsion and lateral bending as follows,

σmax,Ed = My,Ed

Weff,y+ Mfz,Ed

Wfz≤ fy

γM0(11.46)

where Mfz,Ed is the bending moment in the free flange due to the equivalent lateralload as defined in Fig. 10.3 in EN 1993-1-3 (2006), and Wfz is the gross elastic sectionmodulus of the free flange plus 1/5 of the web height for the point of web–flangeintersection, for bending about the z–z axis (see Fig. 11.20). The determination ofMfz,Ed is dependent on the section dimensions, loading position, span length, numberof anti-sag bars, and spring stiffness CD, the detail of which is specified in Section10.1.4.1 in EN 1993-1-3 (2006).

It should be pointed out that Eq. (11.46) applies to only the case where the free flange isunder compression. For the free flange under tension, where due to positive influenceof flange curling and second order effect moment Mfz,Ed may be taken equal to zero.

NEd NEdh

qEd

kh qEd

b

1/5 h 1/6 h

z

z

yy � � 1/5 h

ez1ez2

My,Ed

Weff,y

Mfz,Ed

Wfz

NEd

Aeff

FIGURE 11.20 Superposition of stresses in free flange (copy from EN 1993-1-3(2006) by permission)

Ch11-H5060.tex 12/7/2007 14: 22 page 445


Similar to any compression members, the flexural buckling of the free flange undercompression also need to be considered. The buckling resistance of the free flange isverified by using the following formula

1χLT

My,Ed

Weff,y+ Mfz,Ed

Wfz≤ fy

γM1(11.47)

in which, χLT = 1/[φLT + (φ2LT − 0, 75(λfz)

2)1/2] = reduction factor for flexural buck-

ling of the free flange (χLT ≤ 1,0 and χLT ≤ 1/(λLT)2 )

φLT = 0,51 + αLT(λfz − 0,4

) + 0,75(λfz)2 = factor used to calculate the reduction

factor

where αLT = 0, 34 is the imperfection factor and λfz is the relative slenderness forflexural buckling of the free flange and is determined from

λfz = lfz

ifz

√fyb

π2E(11.48)

where ifz is the radius of gyration of the gross cross-section of the free flange plusthe contributing part of the web for bending about the z–z axis and lfz is the bucklinglength for the free flange which is specified in Section 10.1.4.2 in EN 1993-1-3 (2006).

11.8 WORKING EXAMPLES

In order to assist designers, manufactures of cold-formed steel sections often providedesign manuals for their products, which, in general, include gross section properties,effective section properties, and load tables for a number of specified span lengthsunder various different sets and/or types of connection. The gross section propertiesare calculated based on the geometric dimensions of the section, whereas the effectivesection properties are calculated based on the effective widths by taking into accountthe effects of local and distortion buckling. The load tables can be obtained from eithertests or calculations by considering the following modes of failure:

• Flexural failure involving local and distortional buckling in compression• Lateral–torsional buckling due to insufficient lateral restraints• Excessive deflection• Shear failure• Web crushing under direct loads or reactions• Combined effects between bending and web crushing, and bending and shear.

The load tables design their sections for specific uses and give the performance ofsections under various given circumstances.

The following is an example of calculation of gross and effective section proper-ties for a cold-formed lipped zed section subjected to bending in the plane of web.

Ch11-H5060.tex 12/7/2007 14: 22 page 446


The calculation is based on the method recommended in EN 1993-1-3 (2006). Thedimensions of the cross-section are (where the influence of rounding of the corners isneglected):

Depth of web h = 202 mmWidth of flange in compression bc = 75 mmWidth of flange in tension bt = 75 mmLength of lip c = 20 mmThickness t = 2 mm

The material properties of the section are:

Modulus of elasticity E = 210000 N/mm2

Poisson’s ratio ν = 0,3Basic yield strength fyb = 390 N/mm2

Partial factor γM0 = 1,00

Checking of geometrical proportions

The design method of EN 1993-1-3 (2006) can be applied if the following conditionsare satisfied (Section 5.2):

b/t ≤ 60 bc/t = 75/2 = 37,5 < 60 → okbt/t = 75/2 = 37,5 < 60 → ok

c/t ≤ 50 c/t = 20/2 = 10 < 50 → okh/t ≤ 500 h/t = 202/2 = 101 < 500 → ok

In order to provide sufficient stiffness and avoid primary buckling of the stiffener itself,the size of stiffener should be within the following range (Section 5.2 in EN 1993-1-3(2006)):

0,2 ≤ c/b ≤ 0,6 c/bc = 20/75 = 0,27 → okc/bt = 20/75 = 0,27 → ok

For cold-formed steel sections the section properties are usually calculated based onthe dimensions of the section middle line as follows (see Fig. 11.21),

Depth of web hp = h − t = 202 − 2 = 200 mmWidth of flange in compression bp1 = bc − t = 75 − 2 = 73 mmWidth of flange in tension bp2 = bt − t = 75 − 2 = 73 mmLength of lip cp = c − t/2 = 20 − 2/2 = 19 mm

Calculation of gross section properties

Gross cross-section area:

A = t(2cp + bp1 + bp2 + hp) = 2 × (2 × 19 + 73 + 73 + 200) = 768 mm2

Ch11-H5060.tex 12/7/2007 14: 22 page 447


c p

bp1

bp2

h pt

c p FIGURE 11.21 Symbols used for representing thedimensions of middle lines of a zed section

Position of the neutral axis with regard to the flange in compression:

zc = hp

2= 200

2= 100 mm

Position of the neutral axis with regard to the flange in tension:

zt = hp − zc = 200 − 100 = 100 mm

Second moment of the gross cross-sectional area:

Iy = (h3p + 2c3

p)t

12+ (bp1 + bp2)t3

12+ z2

c bp1t + z2t bp2t + cpt

(zc − cp

2

)2

+ cpt(

zt − cp

2

)2 = 4880000 mm4

Gross section modulus with regard to the flange in compression:

Wy,c = Iy

zc= 4880000

100= 48800 mm3

Gross section modulus with regard to the flange in tension:

Wy,t = Iy

zt= 4880000

100= 48800 mm3

Calculation of effective section properties

The general (iterative) procedure is applied to calculate the effective properties of thecompression flange and the lip (plane element with edge stiffener). The calculationshould be carried out in three steps:

Step 1 Obtain an initial effective cross-section for the stiffener using effective widthsof the flange and lip determined by assuming that the compression flange is doublysupported, the stiffener gives full restraint (K = ∞) and that the design strength is notreduced, that is, σcom,Ed = fyb/γM0.

• Effective width of the compressed flange

Ch11-H5060.tex 12/7/2007 14: 22 page 448


For the internal compression flange the stress ratio ψ = 1 (uniform compression),so the buckling coefficient is taken as kσ = 4. The relative slenderness thus is:

λb,red = λb =√

fyb

σcr,b=√

390570

= 0,827

where σcr,b = π2E12(1 − ν2)

kσ

(bp1/t)2 = 3,142 × 21000012(1 − 0,32)

4(73/2)2 = 570 N/mm2

Since λb,red = 0,827 > 0,673, the width reduction factor for the doubly supportedcompression element is calculated by (Eq. (11.14a))

ρ = λb,red − 0,055(3 + ψ)

λ2b,red

= 0,827 − 0,055(3 + 1)0,8272 = 0,887

The effective width of the compressed flange thus is:

beff = ρbp1 = 0,887 × 73 = 64,8 mm

be1 = be2 = 0,5beff = 0,5 × 64,8 = 32,4 mm

• Effective length of the lipFor the compression lip, the buckling coefficient should be taken as follows(Eq. (11.13d))

kσ = 0,5 if cp/bp1 ≤ 0,35kσ = 0,5 + 0,83(cp/bp1 − 0,35)2/3 if 0,35 ≤ cp/bp1 ≤ 0,6

For cp/bp1 = 19/73 = 0,260 < 0,35 kσ = 0,5.

The relative slenderness is:

λc,red = λc =√

fyb

σcr,c=√

3901051

= 0,609

where σcr,c = π2E12(1 − ν2)

kσ

(cp/t)2 = 3,142 × 21000012(1 − 0,32)

0,5(19/2)2 = 1050 N/mm2

Since λc,red = 0, 609 < 0, 673, the width reduction factor for the outstand compres-sion element thus is given by (Eq. (11.14b)):

ρ = 1,0The effective length of the compression lip thus is:

ceff = ρcp = 1,0 × 19 = 19 mmThe corresponding effective area of the edge stiffener is:

As = t(be2 + ceff ) = 2 × (32,4 + 19) = 103 mm2

Step 2 Use the initial effective cross-section of the stiffener to determine the reductionfactor, allowing for the effects of the distortional buckling. The elastic critical stressof the distortional buckling for the edge stiffener is (Eq. (11.18))

σcr,d = 2√

KEIs

As= 343 N/mm2

Ch11-H5060.tex 12/7/2007 14: 22 page 449


where K = Et3

4(1 − ν2)(b21hp + b3

1)= 0,445 N/mm2

b1 = bp1 − b2e2

2(be2 + ceff )= 62,8 mm

Is = be2t3

12+ c3

eff t12

+ be2t

[c2

eff2(be2 + ceff )

]2

+ ceff t

[ceff

2− c2

eff2(be2 + ceff )

]2

= 3330 mm4

Thickness reduction factor for the edge stiffener is calculated based on the relativeslenderness of the edge stiffener as follows (Eq. (11.19)):

λd =√

fyb

σcr,d=√

390343

= 1,066

χd = 1,0 if λd ≤ 0,65χd = 1,47 − 0,723λd if 0,65 < λd < 1,38χd = 0,66/λd if λd ≥ 1,38

0,65 < λd < 1,38 so χd = 1,47 − 0,723λd = 0,699

Step 3 As the reduction factor for the buckling of the stiffener is χd = 0,699 < 1, iter-ations are required to refine the value of the reduction factor. The iterations arecarried out based on the reduced design strength, σcom,Ed,i = χd,i−1fyb/γM0 to obtainnew effective widths of the lip and flange in the stiffener and recalculate the criticalstress of distortional buckling of the stiffener and thus to obtain new reduction factor.The iteration stops when the reduction factor χd converges. The final values obtainedafter iterations are be2 = 36,1, ceff = 19,0 and χd = 0,689.

• Effective width of the webThe position of the initial neutral axis (web is assumed as fully effective) with regardto the flange in compression is given by

hc =cp

(hp − cp

2

)+ bp2hp + h2

p

2 + c2effχd

2

cp + bp2 + hp + be1 + (be2 + ceff )χd= 106 mm

The stress ratio thus is:

ψ = −hp − hc

hc= −200 − 106

106= −0,89

The corresponding buckling coefficient is calculated by (Eq. (11.13a))

kσ = 7,81 − 6,29ψ + 9,78ψ2 = 21,2

Ch11-H5060.tex 12/7/2007 14: 22 page 450


The relative slenderness thus is:

λh,red = λh =√

fyb

σcr,h=√

3901051

= 0,986

where σcr,h = π2E12(1 − ν2)

kσ

(hp/t)2 = 3,142 × 21000012(1 − 0,32)

21,2(200/2)2 = 402 N/mm2

The width reduction factor thus is (Eq. (11.14a))

ρ = λh,red − 0,055(3 + ψ)

λ2h,red

= 0,986 − 0,055(3 − 0,89)0,9862 = 0,895

The effective width of the zone in compression of the web is:

heff = ρhc = 0,895 × 106 = 94,9 mm

Part of the effective width near the flange is (see Table 11.1):

he1 = 0,4heff = 0,4 × 94,9 = 37,9 mm

Part of the effective width near the neutral axis is:

he2 = 0,6heff = 0,6 × 94,9 = 56,8 mm

Thus,

h1 = he1 = 37,9 mm

h2 = (hp − hc) + he2 = (200 − 106) + 56,8 = 151 mm

The effective widths of the web obtained above are based on the position of theinitial neutral axis (web is assumed as fully effective). To refine the result iterationsare required which is based on the newly obtained effective widths, he1 and he2, todetermine the new position of the neutral axis. The stress ratio, buckling coefficient,relative slenderness, width reduction factor and effective widths of the web thus are re-calculated according to the new position of the neutral axis. Iteration continues until itconverges. The final values obtained after iterations are he1 = 37,9 mm, he2 = 56,8 mmand h2 = 149 mm (see Fig. 11.22).

• Effective properties of the section (see Fig. 11.23)Effective cross-section area:

Aeff = t[cp + bp2 + h2 + h1 + be1 + (be2 + ceff )χd] = 698 mm2

Ch11-H5060.tex 12/7/2007 14: 22 page 451


h e1

h e2

h p

c eff

h c

n.a.

bp1

be1 be2

t

c p

bp2FIGURE 11.22 Symbols used for representingthe dimensions of the effective cross-section

c p

bp2

h 1

h p h e2

h 2

h e1

c eff

t χd

bp1

be1 be2

t

z tz c

n.a

FIGURE 11.23 Symbols used forrepresenting the properties of the effectivecross-section

Position of the neutral axis with regard to the flange in compression:

zc =t[

cp

(hp − cp

2

)+ bp2hp + h2

(hp − h2

2

)+ h2

12 + c2

effχd

2

]Aeff

= 108 mm

Position of the neutral axis with regard to the flange in tension:

zt = hp − zc = 92 mm

Second moment of the effective sectional area:

Ieff,y = t(h31 + h3

2 + c3p + χdc3

eff )

12+ t3(bp2 + be1 + be2χ

3d)

12

+ cpt(

zt − cp

2

)2 + bp2tz2t + h2t

(zt − h2

2

)2

+ h1t(

zc − h1

2

)2

+ be1tz2c + be2(χdt)z2

c + ceff (χdt)(

zc − ceff

2

)2

= 4340000 mm4

Ch11-H5060.tex 12/7/2007 14: 22 page 452


Effective section modulus with regard to the flange in compression:

Weff,y,c = Ieff,y

zc= 40100 mm3

Effective section modulus with regard to the flange in tension:

Weff,y,t = Ieff,y

zt= 47200 mm3

The design value of the resistance of the section to bending moment about the y-axisdue to local and distortional buckling is

Mc,Rd = fybWeff,y,c

γM0= 390 × 40100 × 10−6

1,0= 15,6 kNm

The load table of a section is defined based on the specified type of connection andgiven span lengths. Assume that the section is used as a single span of 5 m long, bothends are butted to rafter beam using cleats, the section is subjected to gravity loadingand is fully restrained laterally.

The design value of the resistance to gravity load on the span due to local anddistortional buckling is:

Pc,Rd = 8Mc,Rd

l= 8 × 15,6

5= 25,0 kN

The design value of the plastic shear resistance is (Section 6.2.6 in EN 1993-1-1 (2005)):

Vpl,Rd =Av

(fyb/

√3)

γM0=

200×2×390√3

× 10−3

1,0= 90,1 kN

The design value of the shear buckling resistance is (Section 6.1.5 in EN 1993-1-3(2006)):

Vb,Rd =hp

sin φtfbv

γM0=

200sin 90◦ × 2 × 0,48×390

1,49 × 10−3

1,0= 50,3 kN

in which,

fbv = 0,58fyb if λw ≤ 0, 83

fbv = 0,48fyb/λw if 0,83 < λw

λw = 0, 346hp

t

√fyb

E= 0,346 × 200

2×√

390210000

= 1,49

So, the design shear resistance is

Vc,Rd = min(Vpl,Rd, Vb,Rd) = 50,3 kN

Since the maximum shear force for the simply supported beam occurs at the support,the value of which is only half of the span load, it is obvious that for the present case

Ch11-H5060.tex 12/7/2007 14: 22 page 453


the design load is controlled by the design value of the resistance due to local anddistortional buckling. Therefore the ultimate design load for the section is 25 kN.

Note that, the load factors used for permanent actions and for variable actions aredifferent. The design load for any combinations of dead and imposed loads thus shouldsatisfy

γGPG + γQPQ ≤ Pc,Rd = 25 kN

where γG = 1,35 and γQ = 1,5 are the load factors for dead and imposed loads, PG andPQ are the dead and imposed loads, respectively.

The deflection check is usually done by using computer software which is normally pro-vided by the manufacturer. The allowable deflection is determined based on individualcases and is not specified in most design standards.

11.9 CHAPTER CONCLUSIONS

This chapter has described the mechanism of failures of the cold-formed steel sectionsand the corresponding methods for analyses and principles for design. The focuseshave been placed on the analyses of local, distortional and lateral–torsional bucklingof the sections for these are the main differences between the cold-formed steel sectionand the hot rolled steel section. For the cold-formed steel sections the most importantproperties are the effective section properties. For summarizing the calculation pro-cedure a flowchart of calculation of effective section properties is given in Figs. 11.24and 11.25.

The design of a cold-formed steel section is much more complicated than that of a hotrolled section. This is partly because the section involves elements which have largewidth-to-thickness ratios and thus are easier to buckle locally and distortionally, andpartly because the section is locally restrained by its supported trapetzoidal sheeting,which complicates the loading system and generates different stresses in restrainedand free flanges. Simplified design methods may be used for channel, zed and sigmapurlin systems that have no anti-sag bars, no use of sleeves and overlapping betweentwo adjacent sections (see Annex E in EN 1993-1-3 (2006)).

It is well known that for a short beam the design is normally controlled by the bendingmoment and/or shear force, while for a long beam it is usually controlled by thedeflection and lateral–torsional buckling. Thus, purlins are usually designed to becontinuous by using sleeves or overlapping in order to satisfy deflection limits andanti-sag bars to prevent twisting during erection and to stabilize the lower flangeagainst wind uplift. For simply supported members, it is the sagging (positive) momentconditions that determine the capacity of the member. For continuous members overone or more internal supports, moments are to be determined elastically. Plastic hingeanalysis is not permitted because the slender sections are not able to maintain their full

Ch11-H5060.tex 12/7/2007 14: 22 page 454


Dimensions ofcross-section;Material data

The influence of rounding of thecorners may neglected if.r/t 5; r/bp 0,1

EN 1993-1-3 §5.1(3)

EN 1993-1-3 §5.2

EN 1993-1-3 §5.5.3.2

Axial compression Major axis bending

EN 1993-1-3 §5.5.3.2

Determine theproperties of effective

cross-section: leff, Weff

Determine the effective section

properties of the web

Determine the effective section

properties of the web

Determine effectivearea of the cross-

section: Aeff

Determine effectivewidths for the flange

and lip in comperssion

Determine effectivewidths for the flanges

and lips incompression

EN 1993-1-3 §5.5.2

EN 1993-1-3 §5.5.2

EN 1993-1-5§4.4

EN 1993-1-5 §4.4

The generai (iterative)procedure is applied tocalculate the effective

properties of the flangesand lips in compression.

The calculation should be carried out in three steps.

Determine geometricalproportions of cross-section

Compute the grosscross-section

properties

No

Yes

The general (iterative)procedure is applied tocalculate the effectiveproperties of the flangeand lip in compression.

The calculation should becarried out in three steps.

Recalculate the positionof the neutral axistaking into account theeffective properties ofcompression flange andlip

χd1, be12,ceff1, be11,χd2, be22,ceff2, be21

he1,he2

Aeff

Start

Stop

StopStop

leff, Weff

he1, he2

χd1,be2, ceff,be1

b/t 60c/t 50

h/t 5000,2 c/b 0,6

?

FIGURE 11.24 The flowchart of calculation of effective section properties forcold-formed steel sections under compression or bending (copy fromwww.access-steel.com by permission)

moment capacity when rotations exceed the point at which the section reaches yield.By utilizing the flexibility of sleeved or overlapping purlins at the supports, some elasticredistribution of moment may be achieved, and hence lead to more efficient design ofthe member. However, when the purlins are sleeved or overlapped, the design of the

Ch11-H5060.tex 12/7/2007 14: 22 page 455

A

For the initial evaluation, it is assumedthat the flange in compression is double

supported (K�∞) and that the design

strength is not reduced (σcom,Ed � fyb / γM0)

Step 1 Determine theeffective cross-section for thestiffener using effective width

of flange

Determine the effective area ofthe edge stiffener: Ae

Step 2 Use the intial effectivecross-section of the stiffener todetermine the reduction factor

Determine the elastic bucklingstress for the edge stiffener.

Determine the reduction factorfor the distortional buckling

resistance of the edge stiffener

Xd < 1 ?

Step 3 iterate to refine thevalue of the reduction factorfor buckling of the stiffener.

σσ,s

Where: K is the spring stiffnessper unit length and Is is theeffective second moment ofarea of the stiffener

The iterations are carried out

based on modified values of ρ,

obtained using: σcom,Ed�fyb / γM0

The iteration stops when the

reduction factor χ converges.

To A

EN 1993-1-3§5.5.3.2(3)

EN 1993-1-3§5.5.3.2(5a)

EN 1993-1-5§4.4

EN 1993-1-3§5.5.3.2(6)

EN 1993-1-3§5.5.3.2(3)

EN 1993-1-3§5.5.3.2(7)

EN 1993-1-3§5.5.3.1(5)

EN 1993-1-3§5.5.3.1(7)

EN 1993-1-5§4.4(2)

EN 1993-1-3§5.5.3.2

(3), (10), (12)

Allow for the effects ofcontinuous spring

restraint

No

Yes

Start

be1,be2,ceff

As

χd

Return

Return

FIGURE 11.25 The flowchart of iterative procedure for calculation of reductionfactor due to distortion buckling (copy from www.access-steel.com by permission)

Ch11-H5060.tex 12/7/2007 14: 22 page 456


system is normally developed based on testing or the combination of analysis andtesting to achieve economic solutions. Standardized testing and evaluation procedureshave been given in EN 1993-1-3 (2006) (Annex A) for both the cold-formed membersand sheeting.

REFERENCES

AS/NZS 4600 (1996). Australia/New Zealand Standard for Cold-formed Steel Structures. StandardsAustralia, Sydney.

Bulson, P.S. (1970). The stability of flat plates. Chatto & Windus, London.Cheung, Y.K. (1976). Finite strip methods in structural analysis. Pergramon Press, New York.Chu, X.T. (2004). Failure Analysis of Cold-formed Steel Sections. Ph.D. Thesis, Aston University,

Birmingham. UK.Chu, X.T., Kettle, R. and Li, L.Y. (2004a). Lateral–torsion buckling analysis of partial-laterally

restrained thin-walled channel-sections beams, Journal of Constructional Steel Research, 60(8),1159–1175.

Chu, X.T., Li, L.Y. and Kettle, R. (2004b). The effect of warping stress on the lateral torsion bucklingof cold-formed zed-purlins, Journal of Applied Mechanics, 71(5), 742–744.

Chu, X.T., Rickard, J. and Li, L.Y. (2005a). Influence of lateral restraint on lateral–torsional bucklingof cold-formed steel purlins. Thin-Walled Structures, 43, 800–810.

Chu, X.T., Ye, Z.M, Kettle, R. and Li, L.Y. (2005b). Buckling behaviour of cold-formed channelsections under uniformly distributed loads. Thin-Walled Structures, 43, 531–542.

Chu, X.T., Ye, Z.M, Li, L.Y. and Kettle, R. (2006). Local and distortional buckling of cold-formed zed-section beams under uniformly distributed transverse loads, International Journal of MechanicalSciences, 48(8), 378–388.

Davies, J.M., Leach, P. and Heinz, D. (1993). Second-order generalized beam theory, Journal ofConstructional Steel Research, 31, 221–241.

EN 1993-1-1 (2005). Eurocode 3: Design of steel structures – Part 1–1: General rules and rules forbuildings. BSI.

EN 1993-1-3 (2006). Eurocode 3: Design of steel structures – Part 1–3: General – Cold formed thin gaugemembers and sheeting. BSI.

EN 1993-1-5 (2006). Eurocode 3: Design of steel structures – Part 1–5: General – Strength and stabilityof planar plated structures without transverse loading. BSI.

Hancock, G.J. (1997). Design for distortional buckling of flexural members, Thin-Walled Structures,27(1), 3–12.

Kesti, J. and Davies, J.M. (1999). Local and distortional buckling of thin-walled short columns,Thin-Walled Structures, 34, 115–134.

Lau, S.C.W. and Hancock, G.J. (1986). Buckling of thin flat-walled structures by a spline finite stripmethod, Thin-Walled Structures, 4, 269–294.

Lau, S.C.W. and Hancock, G.J. (1987). Distortional buckling formulas for channel columns, Journalof Structural Engineering, ASCE, 113(5), 1063–1078.

Lau, S.C.W. and Hancock, G.J. (1989). Inelastic buckling analyses of beams, columns and plates usingthe spline finite strip method, Thin-Walled Structures, 7, 213–238.

Li, L.Y. (2004). Lateral–torsional buckling of cold-formed zed-purlins partial-laterally restrained bymetal sheeting, Thin-Walled Structures, 42(7), 995–1011.

Loughlan, J. (1993). Thin-walled cold-formed sections subjected to compressive loading, Thin-WalledStructures, 16(1–4), 65–109.

Lucas, R.M., Al-Bermani, F.G.A. and Kitipornchai, S. (1997a). Modelling of cold-formed purlin-sheeting systems – Part 1. Full model, Thin-Walled Structures, 27(3), 223–243.

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Lucas, R.M., Al-Bermani, F.G.A. and Kitipornchai, S. (1997b). Modelling of cold-formed purlin-sheeting systems – Part 2. Simplified model, Thin-Walled Structures, 27(4), 263–286.

Oden, J.T. (1967). Mechanics of elastic structures. McGraw-Hill Book Company, New York.Rhodes, J. and Lawson, R.M. (1992). Design of structures using cold-formed steel sections. SCI

Publication, 089. SCI.Schafer, B.W. (1997). Cold-formed Steel Behaviour and Design: Analytical and Numerical Modelling of

Elements and Members with Longitudinal Stiffeners. Ph.D. Thesis. Cornell University.Timoshenko, S.P. and Gere, J.M. (1961). Theory of elastic stability. McGraw-Hill Book Company,

New York.Vlasov, V.Z. (1961). Thin-walled elastic beams. Israel Program for Scientific Translations, Jerusalem,

Israel.Von Karman, T., Sechler, E.E. and Donnell, L.H. (1932). The strength design of thin plates in

compression, Transactions ASME, 54, 53–55.Vrany, T. (2002). Torsional restraint of cold-formed beams provided by corrugated sheeting for

arbitrary input variables. Proceedings of Eurosteel. Coimbra, cmm, 2002, 734–742.Walker, A.C. (1975). Design and analysis of cold-formed sections. International Textbook Company

Ltd, London.Winter, G. (1968). Thin walled structures – theoretical solutions and test results, Preliminary

Publications 8th Congress IABSE, 101–112.Ye, Z.M., Kettle, R., Li, L.Y. and Schafer, B. (2002). Buckling behaviour of cold-formed zed-purlins

partially restrained by steel sheeting, Thin-Walled Structures, 40, 853–864.Ye, Z.M., Kettle, R. and Li, L.Y. (2004). Analysis of cold-formed zed-purlins partially restrained by

steel sheeting, Computer and Structures, 82, 731–739.Zienkiewicz, O.C. and Taylor, R.L. (2000). The finite element method, Butterworth Heinemann,

Oxford.

Annex-H5060.tex 17/7/2007 19: 21 page 458

Annex A1 / Design Strengths forFillet Welds

Design strength per unit length (Fw,Rd)in KN/mm

Steel grade S275 Steel grade S355fu = 430 MPa fu = 510 MPa

Leg length (mm) βw = 0,85 βw = 0,90

4 0,654 0,7335 0,818 0,9166 0,981 1,0998 1,308 1,466

10 1,636 1,83212 1,963 2,19915 2,453 2,74818 2,944 3,29820 3,271 3,66422 3,598 4,03125 4,089 4,580

Notes:(1) Fw,Rd = f u/(31/2 βwγM2) × 0,7 × (leg length) for equal leg lengths and

γM2 = 1,25.(2) Steel to Standard 10025-2 with appropriate electrodes (BSEN 499).

Annex-H5060.tex 17/7/2007 19: 21 page 459

• 459

Annex A2 / Design Strengths forClass 4.6 Ordinary Bolts

Tension Single shear Single shearReduced stress using stress using stress using

Bolt diameter area of bolt reduced area reduced area gross aread As Ft, Rd Fv,Rd Fv,Rd(mm) (mm2) (kN) (kN) (kN)

(M12) 84,3 24,3 16,2 21,7M16 157 45,2 30,1 38,6M20 245 70,6 47,0 60,3(M22) 303 87,3 58,2 73,0M24 353 101,7 67,8 86,9(M27) 459 132,2 88,1 109,9M30 561 161,6 107,7 135,7(M33) 694 199,9 133,2 164,2M36 817 235,3 156,9 195,4

Notes:(1) Fv,Rd = 0,6 (A or As) f ub/γM2(2) F t,Rd = 0,9 Asf ub/γM2(3) f ub = 400 MPa; γM2 = 1,25(4) Bolt sizes in brackets are not preferred.(5) Shear values for other classes of bolt are multiplied by a factor

Class 4.6 4.8 5.6 5.8 6.8 8.8 10.9Factor 1,0 0,83 1,25 1,04 1,25 2,0 2,08

(6) Tension values for other classes of bolt are multiplied by a factorClass 4.6 4.8 5.6 5.8 6.8 8.8 10.9Factor 1,0 1,0 1,25 1,25 1,5 2,0 2,5

Annex-H5060.tex 17/7/2007 19: 21 page 460

460 •Annex A3 / Design Strengths forPreloaded Class 8.8and 10.9 Bolts

Reduced Preload Preload Slip SlipBolt area of force force resistance resistancesize bolt class 8.8 class 10.9 class 8.8 class 10.9d As Fp,C Fp,C F s,Rd F s,Rd(mm) (mm2) (kN) (kN) (kN) (kN)

(M12) 84,3 47,2 59,1 18,9 23,6M16 157 87,9 109,9 35,2 44,0M20 245 137,2 171,5 54,9 68,6M22 303 169,7 212,1 67,9 84,8M24 353 197,7 247,1 79,1 98,8M27 459 257,0 321,3 102,8 128,5M30 561 314,2 392,7 125,7 157,1M36 817 457,5 571,9 183,0 228,8

Notes:(1) Preload Fp,C = 0,7 f ub As where f ub = 800 (Class 8.8) and 1000 (Class 10.9) MPa.(2) Design slip resistance F s,Rd = ksnμFp,C/γM3 where ks = 1, n = 1, μ = 0,5 and

γM3 = 1,25(3) Bolt sizes in brackets are not preferred.

Annex-H5060.tex 17/7/2007 19: 21 page 461

• 461

Annex A4 / Space of Holes in Beams,Columns, Joists and Tees

Beams, columns, joists and tees

Nominal Spacings in millimetres Recommended Actualflange diameter of bolt bminwidths (mm) S1 S2 S3 S4 (mm) (mm)

419–368 140 140 75 290 24 362330–305 140 120 60 240 24 312330–305 140 120 60 240 20 300292–203 140 – – – 24 212190–165 90 – – – 24 162152 90 – – – 20 150146–114 70 – – – 20 130102 54 – – – 12 9889 50 – – – – –76 40 – – – – –64 34 – – – – –51 30 – – – – –

S3 S2

S4

S3

S1S1

S1

Annex-H5060.tex 17/7/2007 19: 21 page 462

462 •

Annex A5 / Spacing of Holes in Angles

Angles

Spacing of holes Maximum diameter of bolt

Nominal S4,S5leg length S1 S2 S3 S4 S5 S6 S1 S2 and S3 and S6(mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm) (mm)

200 – 75 75 55 55 55 – 30 20150 – 55 55 – – – – 20 –125 – 45 50 – – – – 20 –120 – 45 50 – – – – 20 –100 55 – – – – – 24 – –

90 50 – – – – – 24 – –80 45 – – – – – 20 – –75 45 – – – – – 20 – –70 40 – – – – – 20 – –65 35 – – – – – 20 – –60 35 – – – – – 16 – –50 28 – – – – – 12 – –45 25 – – – – – – – –40 23 – – – – – – – –30 20 – – – – – – – –25 15 – – – – – – – –

Leg

Leg

S3

S2

S1

LegS6

S5

S4

Annex-H5060.tex 17/7/2007 19: 21 page 463

• 463

Annex A6 / Spacing of Holes inChannels

Channels

Nominal Recommendedflange width S1 diameter of bolt(mm) (mm) (mm)

102 55 2489 55 2076 45 2064 35 1651 30 1038 22 –

S1

Annex-H5060.tex 17/7/2007 19: 21 page 464

464 • Annex 7 / From BS5950: 1: 1990 by Permission

Annex A7 / From BS5950: 1: 1990by Permission

TABLE 15 Slenderness correction factor, n, for members with applied loading substan-tially concentrated within the middle fifth of the unrestrained length.

Note 1. All hogging moments are +ve.Note 2. β is defined in Table 18.Note 3. M0 is the mid-length momenton a simply supported span equal tothe unrestrained length (see Table 17).

Unrestrained length L

M M0 BM

L10 maximum

β positive β negative

γ = M/M0 1.0 0.8 0.6 0.4 0.2 0.0 −0.2 −0.4 −0.6 −0.8 −1.0

+50.00 1.00 0.96 0.92 0.87 0.82 0.77 0.72 0.67 0.66 0.66 0.65+10.00 0.99 0.99 0.94 0.90 0.85 0.80 0.75 0.69 0.68 0.68 0.67+5.00 0.98 0.98 0.97 0.93 0.89 0.84 0.79 0.73 0.71 0.70 0.70+2.00 0.96 0.95 0.95 0.95 0.94 0.94 0.89 0.84 0.79 0.77 0.76+1.50 0.95 0.95 0.94 0.94 0.93 0.93 0.92 0.90 0.85 0.80 0.80+1.00 0.93 0.92 0.92 0.92 0.92 0.91 0.91 0.91 0.91 0.92 0.92+0.50 0.90 0.90 0.90 0.89 0.89 0.89 0.89 0.89 0.88 0.88 0.88

0.00 0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86−0.10 0.85 0.85 0.85 0.85 0.85 0.86 0.86 0.86 0.86 0.86 0.86−0.20 0.83 0.83 0.83 0.84 0.84 0.85 0.85 0.85 0.86 0.86 0.86−0.30 0.81 0.82 0.82 0.83 0.83 0.84 0.85 0.85 0.86 0.86 0.87−0.40 0.79 0.80 0.81 0.81 0.82 0.83 0.84 0.85 0.85 0.86 0.87−0.50 0.77 0.78 0.79 0.80 0.82 0.83 0.85 0.86 0.86 0.87 0.88−0.60 0.62 0.66 0.72 0.77 0.80 0.82 0.84 0.85 0.86 0.87 0.88−0.70 0.56 0.56 0.61 0.67 0.73 0.79 0.83 0.85 0.87 0.88 0.89−0.80 0.56 0.53 0.54 0.59 0.65 0.71 0.77 0.83 0.89 0.90 0.90−0.90 0.59 0.57 0.54 0.53 0.57 0.64 0.71 0.77 0.84 0.88 0.91−1.00 0.62 0.58 0.54 0.52 0.54 0.59 0.66 0.72 0.80 0.85 0.92

(Continued)

Annex-H5060.tex 17/7/2007 19: 21 page 465


TABLE 15 Continued

−1.10 0.66 0.62 0.57 0.54 0.54 0.57 0.63 0.68 0.76 0.83 0.89−1.20 0.70 0.66 0.60 0.55 0.54 0.55 0.60 0.65 0.73 0.80 0.87−1.30 0.73 0.69 0.63 0.57 0.55 0.54 0.57 0.61 0.69 0.77 0.83−1.40 0.74 0.70 0.64 0.58 0.56 0.54 0.55 0.60 0.66 0.74 0.81−1.50 0.75 0.70 0.64 0.59 0.56 0.54 0.55 0.59 0.65 0.73 0.80−1.60 0.76 0.72 0.65 0.60 0.57 0.55 0.55 0.58 0.64 0.72 0.80−1.70 0.77 0.74 0.66 0.61 0.58 0.56 0.55 0.58 0.63 0.70 0.78−1.80 0.79 0.77 0.68 0.63 0.59 0.56 0.56 0.57 0.62 0.69 0.76−1.90 0.80 0.79 0.69 0.64 0.60 0.57 0.56 0.57 0.61 0.67 0.75−2.00 0.81 0.81 0.70 0.65 0.61 0.58 0.56 0.56 0.60 0.66 0.74−5.00 0.93 0.89 0.83 0.77 0.72 0.67 0.64 0.61 0.60 0.62 0.65−50.00 0.99 0.95 0.90 0.86 0.79 0.74 0.70 0.67 0.64 0.63 0.65Infinity 1.00 0.96 0.91 0.86 0.82 0.77 0.72 0.68 0.65 0.65 0.65

Note 4. The values of n in this table apply only to members of UNIFORM section.Note 5. Values for intermediate values of β and γ may be interpolated.

TABLE 16 Slenderness correction factor, n, for members with applied loading other thanas for Table 15.

Note 1. All hogging moments are +ve.Note 2. β is defined in Table 18.Note 3. M0 is the mid-length momenton a simply supported span equal tothe unrestrained length (see Table 17).

Unrestrained length L

M M0 BM

β positive β negative

γ = M/M0 1.0 0.8 0.6 0.4 0.2 0.0 −0.2 −0.4 −0.6 −0.8 −1.0

+50.00 1.00 0.96 0.92 0.87 0.83 0.77 0.72 0.67 0.66 0.66 0.65+10.00 0.99 0.98 0.95 0.91 0.86 0.81 0.76 0.70 0.68 0.68 0.67+5.00 0.99 0.98 0.97 0.94 0.90 0.85 0.80 0.75 0.71 0.70 0.70+2.00 0.98 0.98 0.97 0.96 0.94 0.92 0.90 0.86 0.82 0.78 0.76+1.50 0.97 0.97 0.97 0.96 0.95 0.93 0.92 0.89 0.86 0.83 0.79+1.00 0.97 0.97 0.97 0.96 0.96 0.95 0.94 0.93 0.93 0.91 0.89+0.50 0.96 0.96 0.96 0.96 0.96 0.95 0.94 0.94 0.94 0.93 0.92

0.00 0.94 0.94 0.94 0.94 0.94 0.94 0.94 0.94 0.94 0.94 0.94

(Continued)

Annex-H5060.tex 17/7/2007 19: 21 page 466

466 • Annex 7 / From BS5950: 1: 1990 by Permission

TABLE 16 Continued

−0.10 0.93 0.93 0.93 0.93 0.94 0.94 0.94 0.94 0.94 0.94 0.94−0.20 0.92 0.92 0.92 0.92 0.93 0.93 0.93 0.93 0.94 0.94 0.93−0.30 0.91 0.91 0.92 0.92 0.93 0.93 0.93 0.93 0.94 0.94 0.94−0.40 0.90 0.90 0.91 0.91 0.92 0.92 0.92 0.92 0.93 0.93 0.93−0.50 0.89 0.90 0.91 0.91 0.92 0.92 0.92 0.92 0.92 0.92 0.92−0.60 0.71 0.77 0.84 0.87 0.89 0.91 0.92 0.92 0.92 0.92 0.92−0.70 0.57 0.64 0.70 0.77 0.82 0.87 0.89 0.91 0.92 0.92 0.91−0.80 0.47 0.52 0.59 0.67 0.73 0.80 0.86 0.90 0.92 0.92 0.92−0.90 0.47 0.46 0.50 0.58 0.65 0.73 0.80 0.87 0.90 0.90 0.90−1.00 0.50 0.48 0.46 0.51 0.58 0.66 0.73 0.81 0.87 0.89 0.89−1.10 0.54 0.51 0.48 0.49 0.54 0.61 0.69 0.77 0.83 0.87 0.88−1.20 0.57 0.54 0.50 0.47 0.51 0.56 0.64 0.73 0.80 0.84 0.87−1.30 0.61 0.56 0.52 0.47 0.49 0.53 0.61 0.70 0.77 0.82 0.86−1.40 0.64 0.59 0.55 0.49 0.48 0.51 0.58 0.67 0.74 0.79 0.85−1.50 0.67 0.62 0.57 0.51 0.47 0.49 0.56 0.64 0.71 0.77 0.84−1.60 0.69 0.64 0.59 0.52 0.48 0.50 0.55 0.63 0.69 0.76 0.83−1.70 0.71 0.66 0.60 0.54 0.50 0.51 0.55 0.61 0.68 0.74 0.82−1.80 0.74 0.69 0.62 0.55 0.51 0.51 0.54 0.60 0.66 0.73 0.81−1.90 0.76 0.71 0.63 0.57 0.53 0.52 0.54 0.58 0.65 0.71 0.80−2.00 0.78 0.73 0.65 0.58 0.54 0.53 0.53 0.57 0.63 0.70 0.79−5.00 0.91 0.86 0.80 0.74 0.70 0.65 0.62 0.59 0.58 0.61 0.67−50.00 0.99 0.95 0.89 0.84 0.79 0.74 0.70 0.66 0.63 0.62 0.65Infinity 1.00 0.96 0.91 0.86 0.82 0.77 0.72 0.68 0.65 0.65 0.65

Note 4. The values of n in this table apply only to members of UNIFORM section.Note 5. Values for intermediate values of β and γ may be interpolated.

Annex-H5060.tex 17/7/2007 19: 21 page 467


TABLE 17 Moment diagrambetween adjacent points oflateral restraint.

M0

M BM

b �ve g �ve

M0

M BM

b �ve g �ve

M0

M BM

b �ve g �ve

M0

M

BMb�ve g�ve

M0

M

BMb�ve g�ve

M0

M

BM

b�ve g�ve

Index-H5060.tex 11/7/2007 14: 14 page 469

Index

Actionsaccidental, 29, 287, 294characteristic, 27classification, 27combinations, 30, 287design values, 29destabilizing, 98, 104envelopes, 32erection, 29, 301, 359fundamental, 30gravity, 286pattern loading, 31, 103permanent, 28progressive collapse, 295seismic, 30, 287snow, 28, 286stabilising, 104thermal, 29transmission, 298variable, 28wind, 29, 286

analysis of frameselastic, 309pitched roof portal, 314plastic, 309

angles and teesas beams, 42as tension members, 175

Beamsdeflection, 49elastic analysis, 38laterally restrained, 35, 37principal axes, 39section axes, 39shear stress, 54span/depth ratios, 53symmetrical, 41transverse forces, 78types, 36unsymmetrical, 42

boltsbearing strength, 213clearance, 209close tolerance, 208design tables ordinary, Annex A2design tables preloaded, Annex A3edge and end distances, 210foundation, 208holes, 209

large grip lengths, 216long joints, 216net area, 212ordinary bolts,packings, 216preferred sizes, 207prying forces, 218shear strength and deformation, 214shear strength and tension, 213sizes, 213slip resistant, 217spacing, 210tension and shear, 213types, 207washers, 209

bracing, 288, 300brittle fracture, 22buckling

flange, 424gusset plates, 223lateral torsional buckling, 35, 91, 430lip, 424local, 35, 419Perry-robertson, 183struts, 180theory, 181webs, 138

characteristic load, 27characteristic strength, 18Charpy V-notch test, 22classification of sections, 4, 371cold formed sections

analytical model, 414deflections, 434distortional buckling, 424effective width, 422effective section properties, 454elastic local buckling stress, 420finite strip methods, 435lateral torsional buckling, 430local buckling, 419post buckling behaviour, 422

columns, 181, 300, 312, 325composite constructioncomposite beams

deflection, 382elastic capacity, 376flexural design, 371flexural shear, 377

Index-H5060.tex 11/7/2007 14: 14 page 470

470 • Index

composite beams (continued)longitudinal shear, 379partial shear connection, 376, 378shear connector design, 377transverse shear, 380vibration, 382

composite columnsaxial compression, 396biaxial bending, 397buckling, 401design moments, 401determination of member capacity, 398fire, 402uniaxial bending, 396

composite slabsdeflection, 360, 362flexural shear, 361, 362flexure, 360, 361longitudinal shear, 362, 379punching shear, 362shear bond, 360transverse shear, 380

compression membersaxially loaded, 175buckling length, 186, 312buckling theory, 181compression members, 180effective length, 185local buckling, 185slenderness ratio, 185with moments, 190, 401

connections (see joints)corrosion of steel, 20

Dead loads (see actions)deflections

limits, 53beams, 49composite slabs, 360portal frame, 325span/depth ratios, 53

deformations, 287design

envelope, 19methods for frames, 198rigid, 294, 39, 311sem-rigid, 294, 309, 310simple, 309, 310strength, 19

ductility, 20durability, 20drawings, 12

Effective area of an angle tie, 175effective length

beam, 97cantilever, 100, 105column, 185, 312

errors in design, 11expansion and contraction joints, 304

Fabrication of steelwork, 12failure criteria

bolts, 213steel, 24welds, 203

fatigue, 23fire

single storey structures, 302multi-storey structures, 302composite columns, 402

flooring systems, 284frames

classification, 306imperfections, 308single storey, 282, 288multi-storey, 283, 293

forcesbracing, 290, 300erection, 301seismic, 287tying, 296

foundation design 303frame types, 2

Gable end frame, 336gusset plates, 223, 250, 265

Hardness, 20, 317haunches, 314, 317, 332holes in members, Annexes A4, A5, A6hollow sections, 234, 269

lateral torsional buckling, 123hydrogen cracks, 205

Intermediate vertical stiffener, 145

Jointsbeam splice, 233, 257beam-to-beam, 243, 256beam-to-column, 232, 242, 252column bracket, 231, 237, 247column-to-column, 233, 260column-to-foundation, 234, 245, 263construction movement, 304deformation, 205eccentric shear, 225effective length, 202end bearing, 226effective length, 202expansion and contraction, 304fillet weld, 203ideal, 199knee for portal frame, 235, 271long 202‘pinned’, 228RHS-to-RHS, 234, 269‘rigid’, 230roof truss, 235rotational stiffness, 275washers, 209welded, 199

Index-H5060.tex 11/7/2007 14: 14 page 471

Index • 471

Lateralrestraint, 77torsional buckling, 36, 91, 123, 125

limit state design, 9lintels, 299loads (see actions)

Manufacture of steel, 5material variation, 17multi-storey frames, 283, 293

Overhanging beams, 107

Partial safety factors, 18plastic analysis

effect of axial load, 177effect of shear force, 73hinges, 37lateral restraint, 77methods of analysis, 71reduction of modulus due to shear, 73, 376section properties, 67theorems, 72

plate girderfabrication, 132intermediate vertical stiffeners, 145lateral torsional buckling, 134load bearing stiffeners, 142minimum web thickness, 133minimum weight, 134web buckling, 138welding, 132, 146

plate thicknesselastic method, 222plastic method, 218standard thicknesses, 3

Poisson’s ratio, 19portal frame

single bay, 314multi-bay, 337

principal axes, 38progressive collapse, 295prying force, 218

Rafter stability, 332residual stresses, 22, 94restraints, 99, 300, 325, 332

Sectionaxes, 39cold formed, 413classification, 35elastic properties, 38modulus, 41second moment of area, 40types, 3unsymmetrical, 42plastic properties, 67

seismic actions, 287settlement, 304

shearcentre, 57modulus, 19closed sections, 54lag, 57open sections, 56plate grider, 138stress, 54

sign conventions, 39single storey frames, 282, 288slenderness ratio, 185splices

column, 300plate, 146

stability of structures, 296, 305steel

cold formed, 3first use, 5manufacture, 5production, 6standard sections, 3stock sizes, 3types of, 3

stiffenersload bearing, 142plate girder, 142transverse, 144

strengthcharacteristic, 18design, 19

stressconcentrations, 24residual, 22strain relationship, 38yield, 38

structural design, 7structural framing, 10

T-stubs, 218torsional buckling interaction, 131tension members

angles, 175axially loaded, 175with moments, 177net area, 210effective area, 176

thermal coefficient of expansion, 19testing of steel, 14tolerances, 12torsional buckling, 127torsional restraint, 62, 97, 100toughness, 22transverse forces, 78trusses

non-triangulated trusses, 347triangulated, 341

Vibration, 382vierendeel girder, 348

Index-H5060.tex 11/7/2007 14: 14 page 472

472 • Index

Washers, 209web buckling of plate girders, 138welds

brittle fracture, 206butt welds, 201defects, 205design table, Annex A1effective length, 202electrodes, 199factors affecting strength, 205failure criteria, 204hydrogen cracks, 206

lamellar tearing, 206load-deformation, 205stress concentrations, 206throat thickness, 202types, 201

windbracing, 288girder, 291loading, 286

Yield stress, 18Young’s modulus, 19

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